4.2: Infinite Series
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Learning Objectives
- Explain the meaning of the sum of an infinite series.
- Calculate the sum of a geometric series.
- Evaluate a telescoping series.
We have seen that a sequence is an ordered set of terms. If you add these terms together, you get a series. In this section we define an infinite series and show how series are related to sequences. We also define what it means for a series to converge or diverge. We introduce one of the most important types of series: the geometric series. We will use geometric series in the next chapter to write certain functions as polynomials with an infinite number of terms. This process is important because it allows us to evaluate, differentiate, and integrate complicated functions by using polynomials that are easier to handle. We also discuss the harmonic series, arguably the most interesting divergent series because it just fails to converge.
Sums and Series
An infinite series is a sum of infinitely many terms and is written in the form
∞∑n=1an=a1+a2+a3+⋯.
But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the limit of partial sums. A partial sum of an infinite series is a finite sum of the form
k∑n=1an=a1+a2+a3+⋯+ak.
To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that 1000 gallons enters the lake the first week. During the second week, an additional 500 gallons of oil enters the lake. The third week, 250 more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after k weeks. Letting Sk denote the amount of oil in the lake (measured in thousands of gallons) after k weeks, we see that
S1=1
S2=1+0.5=1+12
S3=1+0.5+0.25=1+12+14
S4=1+0.5+0.25+0.125=1+12+14+18
S5=1+0.5+0.25+0.125+0.0625=1+12+14+18+116.
Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after k weeks is
Sk=1+12+14+18+116+⋯+12k−1=k∑n=1(12)n−1.
We are interested in what happens as k→∞. Symbolically, the amount of oil in the lake as k→∞ is given by the infinite series
∞∑n=1(12)n−1=1+12+14+18+116+⋯.
At the same time, as k→∞, the amount of oil in the lake can be calculated by evaluating limk→∞Sk. Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums Sk. If the sequence of partial sums Sk converges, we say that the infinite series converges, and its sum is given by limk→∞Sk. If the sequence Sk diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence Sk.
First, simplifying some of these partial sums, we see that
S1=1
S2=1+12=32
S3=1+12+14=74
S4=1+12+14+18=158
S5=1+12+14+18+116=3116.
Plotting some of these values in Figure 4.2.1, it appears that the sequence Sk could be approaching 2.

Let’s look for more convincing evidence. In the following table, we list the values of Sk for several values of k.
k | 5 | 10 | 15 | 20 |
---|---|---|---|---|
Sk | 1.9375 | 1.998 | 1.999939 | 1.999998 |
These data supply more evidence suggesting that the sequence Sk converges to 2. Later we will provide an analytic argument that can be used to prove that limk→∞Sk=2. For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to 2. Since this sequence of partial sums converges to 2, we say the infinite series converges to 2 and write
∞∑n=1(12)n−1=2.
Returning to the question about the oil in the lake, since this infinite series converges to 2, we conclude that the amount of oil in the lake will get arbitrarily close to 2000 gallons as the amount of time gets sufficiently large.
This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.
Definition
An infinite series is an expression of the form
∞∑n=1an=a1+a2+a3+⋯.
For each positive integer k, the sum
Sk=k∑n=1an=a1+a2+a3+⋯+ak
is called the kth partial sum of the infinite series. The partial sums form a sequence Sk. If the sequence of partial sums converges to a real number S, the infinite series converges. If we can describe the convergence of a series to S, we call S the sum of the series, and we write
∞∑n=1an=S.
If the sequence of partial sums diverges, we have the divergence of a series.
Note that the index for a series need not begin with n=1 but can begin with any value. For example, the series
∞∑n=1(12)n−1
can also be written as
∞∑n=0(12)nor∞∑n=5(12)n−5.
Often it is convenient for the index to begin at 1, so if for some reason it begins at a different value, we can re-index by making a change of variables. For example, consider the series
∞∑n=21n2.
By introducing the variable m=n−1, so that n=m+1, we can rewrite the series as
∞∑m=11(m+1)2.
Example 4.2.1: Evaluating Limits of Sequences of Partial Sums
For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.
- ∞∑n=1nn+1
- ∞∑n=1(−1)n
- ∞∑n=11n(n+1)
Solution
a. The sequence of partial sums Sk satisfies
S1=12
S2=12+23
S3=12+23+34
S4=12+23+34+45.
Notice that each term added is greater than 1/2. As a result, we see that
S1=12
S2=12+23>12+12=2(12)
S3=12+23+34>12+12+12=3(12)
S4=12+23+34+45>12+12+12+12=4(12).
From this pattern we can see that Sk>k(12) for every integer k. Therefore, Sk is unbounded and consequently, diverges. Therefore, the infinite series ∞∑n=1nn+1 diverges.
b. The sequence of partial sums Sk satisfies
S1=−1
S2=−1+1=0
S3=−1+1−1=−1
S4=−1+1−1+1=0.
From this pattern we can see the sequence of partial sums is
Sk=−1,0,−1,0,….
Since this sequence diverges, the infinite series ∞∑n=1(−1)n diverges.
c. The sequence of partial sums Sk satisfies
S1=11⋅2=12
S2=11⋅2+12⋅3=12+16=23
S3=11⋅2+12⋅3+13⋅4=12+16+112=34
S4=11⋅2+12⋅3+13⋅4+14⋅5=45
S5=11⋅2+12⋅3+13⋅4+14⋅5+15⋅6=56.
From this pattern, we can see that the kth partial sum is given by the explicit formula
Sk=kk+1.
Since k/(k+1)→1, we conclude that the sequence of partial sums converges, and therefore the infinite series converges to 1. We have
∞∑n=11n(n+1)=1.
Exercise 4.2.1
Determine whether the series ∞∑n=1n+1n converges or diverges.
- Hint
-
Look at the sequence of partial sums.
- Answer
-
The series diverges because the kth partial sum Sk>k.
The Harmonic Series
A useful series to know about is the harmonic series. The harmonic series is defined as
∞∑n=11n=1+12+13+14+⋯.
This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums Sk approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence Sk in the following table.
k | 10 | 100 | 1000 | 10,00 | 100,000 | 1,000,000 |
---|---|---|---|---|---|---|
Sk | 2.92897 | 5.18738 | 7.48547 | 9.78761 | 12.09015 | 14.39273 |
Even after 1,000,000 terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.
To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:
S1=1
S2=1+12
S3=1+12+13
S4=1+12+13+14.
Notice that for the last two terms in S4,
13+14>14+14
Therefore, we conclude that
S4>1+12+(14+14)=1+12+12=1+2(12).
Using the same idea for S8, we see that
S8=1+12+13+14+15+16+17+18>1+12+(14+14)+(18+18+18+18)=1+12+12+12=1+3(12).
From this pattern, we see that S1=1,S2=1+1/2,S4>1+2(1/2), and S8>1+3(1/2). More generally, it can be shown that S2j>1+j(1/2) for all j>1. Since 1+j(1/2)→∞, we conclude that the sequence Sk is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since Sk is unbounded, it diverges. Thus, the harmonic series diverges.
Algebraic Properties of Convergent Series
Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.
Note 4.2.1: Algebraic Properties of Convergent Series
Let ∞∑n=1an and ∞∑n=1bn be convergent series. Then the following algebraic properties hold.
i. The series ∞∑n=1(an+bn) converges, and ∞∑n=1(an+bn)=∞∑n=1an+∞∑n=1bn. (Sum Rule)
ii. The series ∞∑n=1(an−bn) converges, and ∞∑n=1(an−bn)=∞∑n=1an−∞∑n=1bn. (Difference Rule)
iii. For any real number c, the series ∞∑n=1can converges, and ∞∑n=1can=c∞∑n=1an. (Constant Multiple Rule)
Example 4.2.2: Using Algebraic Properties of Convergent Series
Evaluate ∞∑n=1[3n(n+1)+(12)n−2].
Solution
We showed earlier that
∞∑n=11n(n+1)=1
and
∞∑n=1(12)n−1=2.
Since both of those series converge, we can apply the properties of Note 4.2.1 to evaluate
∞∑n=1[3n(n+1)+(12)n−2].
Using the sum rule, write
∞∑n=1[3n(n+1)+(12)n−2]=∞∑n=13n(n+1)+∞∑n=1(12)n−2.
Then, using the constant multiple rule and the sums above, we can conclude that
∞∑n=13n(n+1)+∞∑n=1(12)n−2=3∞∑n=11n(n+1)+(12)−1∞∑n=1(12)n−1=3(1)+(12)−1(2)=3+2(2)=7.
Exercise 4.2.2
Evaluate ∞∑n=152n−1.
- Hint
-
Rewrite as ∞∑n=15(12)n−1.
- Answer
-
10
Geometric Series
A geometric series is any series that we can write in the form
a+ar+ar2+ar3+⋯=∞∑n=1arn−1.
Because the ratio of each term in this series to the previous term is r, the number r is called the ratio. We refer to a as the initial term because it is the first term in the series. For example, the series
∞∑n=1(12)n−1=1+12+14+18+⋯
is a geometric series with initial term a=1 and ratio r=1/2.
In general, when does a geometric series converge? Consider the geometric series
∞∑n=1arn−1
when a>0. Its sequence of partial sums Sk is given by
Sk=k∑n=1arn−1=a+ar+ar2+⋯+ark−1.
Consider the case when r=1. In that case,
Sk=a+a(1)+a(1)2+⋯+a(1)k−1=ak.
Since a>0, we know ak→∞ as k→∞. Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for r=1. For r≠1, to find the limit of Sk, multiply Equation by 1−r. Doing so, we see that
(1−r)Sk=a(1−r)(1+r+r2+r3+⋯+rk−1)=a[(1+r+r2+r3+⋯+rk−1)−(r+r2+r3+⋯+rk)]=a(1−rk).
All the other terms cancel out.
Therefore,
Sk=a(1−rk)1−r for r≠1.
From our discussion in the previous section, we know that the geometric sequence rk→0 if |r|<1 and that rk diverges if |r|>1 or r=±1. Therefore, for |r|<1,Sk→a1−r and we have
∞∑n=1arn−1=a1−rif|r|<1.
If |r|≥1,Sk diverges, and therefore
∞∑n=1arn−1diverges if|r|≥1.
Definitions: Diverging and Converging Series
A geometric series is a series of the form
∞∑n=1arn−1=a+ar+ar2+ar3+⋯.
If |r|<1, the series converges, and
∞∑n=1arn−1=a1−rfor|r|<1.
If |r|≥1, the series diverges.
Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than n=1 or the exponent involves a linear expression for n other than n−1. As long as we can rewrite the series in the form given by Equation ???, it is a geometric series. For example, consider the series
∞∑n=0(23)n+2.
To see that this is a geometric series, we write out the first several terms:
∞∑n=0(23)n+2=(23)2+(23)3+(23)4+⋯=49+49⋅(23)+49⋅(23)2+⋯.
We see that the initial term is a=4/9 and the ratio is r=2/3. Therefore, the series can be written as
∞∑n=149⋅(23)n−1.
Since r=2/3<1, this series converges, and its sum is given by
∞∑n=149⋅(23)n−1=4/91−2/3=43.
Example 4.2.3: Determining Convergence or Divergence of a Geometric Series
Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.
- ∞∑n=1(−3)n+14n−1
- ∞∑n=1e2n
Solution
a. Writing out the first several terms in the series, we have
∞∑n=1(−3)n+14n−1=(−3)240+(−3)34+(−3)442+⋯=(−3)2+(−3)2⋅(−34)+(−3)2⋅(−34)2+⋯=9+9⋅(−34)+9⋅(−34)2+⋯.
The initial term a=−3 and the ratio r=−3/4. Since |r|=3/4<1, the series converges to
91−(−3/4)=97/4=367.
b. Writing this series as
e2∞∑n=1(e2)n−1
we can see that this is a geometric series where r=e2>1. Therefore, the series diverges.
Exercise 4.2.3
Determine whether the series ∞∑n=1(−25)n−1 converges or diverges. If it converges, find its sum.
- Hint
-
r=−2/5
- Answer
-
5/7
We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.
Example 4.2.4: Writing Repeating Decimals as Fractions of Integers
Use a geometric series to write 3.¯26 as a fraction of integers.
Solution
Since 3.¯26—=3.262626…, first we write
3.262626…=3+26100+2610,000+26100,000+⋯=3+26102+26104+26106+⋯.
Ignoring the term 3, the rest of this expression is a geometric series with initial term a=26/102 and ratio r=1/102. Therefore, the sum of this series is
26/1021−(1/102)=26/10299/102=2699.
Thus,
3.262626…=3+2699=32399.
Exercise 4.2.4
Write 5.2ˉ7 as a fraction of integers.
- Hint
-
By expressing this number as a series, find a geometric series with initial term a=7/100 and ratio r=1/10.
- Answer
-
475/90
Example 4.2.5: Finding the Area of the Koch Snowflake
Define a sequence of figures {Fn} recursively as follows (Figure 4.2.2). Let F0 be an equilateral triangle with sides of length 1. For n≥1, let Fn be the curve created by removing the middle third of each side of Fn−1 and replacing it with an equilateral triangle pointing outward. The limiting figure as n→∞ is known as Koch’s snowflake.

- Find the length Ln of the perimeter of Fn. Evaluate limn→∞Ln to find the length of the perimeter of Koch’s snowflake.
- Find the area An of figure Fn. Evaluate limn→∞An to find the area of Koch’s snowflake.
Solution
a. Let Nn denote the number of sides of figure Fn. Since F0 is a triangle, N0=3. Let ln denote the length of each side of Fn. Since F0 is an equilateral triangle with sides of length l0=1, we now need to determine N1 and l1. Since F1 is created by removing the middle third of each side and replacing that line segment with two line segments, for each side of F0, we get four sides in F1. Therefore, the number of sides for F1 is
N1=4⋅3.
Since the length of each of these new line segments is 1/3 the length of the line segments in F0, the length of the line segments for F1 is given by
l1=13⋅1=13.
Similarly, for F2, since the middle third of each side of F1 is removed and replaced with two line segments, the number of sides in F2 is given by
N2=4N1=4(4⋅3)=42⋅3.
Since the length of each of these sides is 1/3 the length of the sides of F1, the length of each side of figure F2 is given by
l2=13⋅l1=13⋅13=(13)2.
More generally, since Fn is created by removing the middle third of each side of Fn−1 and replacing that line segment with two line segments of length 13ln−1 in the shape of an equilateral triangle, we know that Nn=4Nn−1 and ln=ln−13. Therefore, the number of sides of figure Fn is
Nn=4n⋅3
and the length of each side is
ln=(13)n.
Therefore, to calculate the perimeter of Fn, we multiply the number of sides Nn and the length of each side ln. We conclude that the perimeter of Fn is given by
Ln=Nn⋅ln=3⋅(43)n
Therefore, the length of the perimeter of Koch’s snowflake is
L=limn→∞Ln=∞.
b. Let Tn denote the area of each new triangle created when forming Fn. For n=0,T0 is the area of the original equilateral triangle. Therefore, T0=A0=√3/4. For n≥1, since the lengths of the sides of the new triangle are 1/3 the length of the sides of Fn−1, we have
Tn=(13)2⋅Tn−1=19⋅Tn−1.
Therefore, Tn=(19)n⋅√34. Since a new triangle is formed on each side of Fn−1,
An=An−1+Nn−1⋅Tn=An−1+(3⋅4n−1)⋅(19)n⋅√34=An−1+34⋅(49)n⋅√34.
Writing out the first few terms A0,A1,A2, we see that
A0=√34
A1=A0+34⋅(49)⋅√34=√34+34⋅(49)⋅√34=√34[1+34⋅(49)]
A2=A1+34⋅(49)2⋅√34=√34[1+34⋅(49)]+34⋅(49)2⋅√34=√34[1+34⋅(49)+34⋅(49)2].
More generally,
An=√34[1+34(49+(49)2+⋯+(49)n)].
Factoring 4/9 out of each term inside the inner parentheses, we rewrite our expression as
An=√34[1+13(1+49+(49)2+⋯+(49)n−1)].
The expression 1+(49)+(49)2+⋯+(49)n−1 is a geometric sum. As shown earlier, this sum satisfies
1+49+(49)2+⋯+(49)n−1=1−(4/9)n1−(4/9).
Substituting this expression into the expression above and simplifying, we conclude that
An=√34[1+13(1−(4/9)n1−(4/9))]=√34[85−35(49)n].
Therefore, the area of Koch’s snowflake is
A=limn→∞An=2√35.
Analysis
The Koch snowflake is interesting because it has finite area, yet infinite perimeter. Although at first this may seem impossible, recall that you have seen similar examples earlier in the text. For example, consider the region bounded by the curve y=1/x2 and the x-axis on the interval [1,∞). Since the improper integral
∫∞11x2dx
converges, the area of this region is finite, even though the perimeter is infinite.
Telescoping Series
Consider the series ∞∑n=11n(n+1). We discussed this series in Example 4.2.1c, showing that the series converges by writing out the first several partial sums S1,S2,…,S6 and noticing that they are all of the form Sk=kk+1. Here we use a different technique to show that this series converges. By using partial fractions, we can write
1n(n+1)=1n−1n+1.
Therefore, the series can be written as
∞∑n=1[1n−1n+1]=(1−12)+(12−13)+(13−14)+⋯.
Writing out the first several terms in the sequence of partial sums Sk, we see that
S1=1−12
S2=(1−12)+(12−13)=1−13
S3=(1−12)+(12−13)+(13−14)=1−14.
In general,
Sk=(1−12)+(12−13)+(13−14)+⋯+(1k−1k+1)=1−1k+1.
We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since Sk=1−1/(k+1) and 1/(k+1)→0 as k→∞, the sequence of partial sums converges to 1, and therefore the series converges to 1.
Definition
A telescoping series is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.
For example, any series of the form
∞∑n=1[bn−bn+1]=(b1−b2)+(b2−b3)+(b3−b4)+⋯
is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that
S1=b1−b2
S2=(b1−b2)+(b2−b3)=b1−b3
S3=(b1−b2)+(b2−b3)+(b3−b4)=b1−b4.
In general, the kth partial sum of this series is
Sk=b1−bk+1.
Since the kth partial sum can be simplified to the difference of these two terms, the sequence of partial sums Sk will converge if and only if the sequence bk+1 converges. Moreover, if the sequence bk+1 converges to some finite number B, then the sequence of partial sums converges to b1−B, and therefore
∞∑n=1[bn−bn+1]=b1−B.
In the next example, we show how to use these ideas to analyze a telescoping series of this form.
Example 4.2.6: Evaluating a Telescoping Series
Determine whether the telescoping series
∞∑n=1[cos(1n)−cos(1n+1)]
converges or diverges. If it converges, find its sum.
Solution
By writing out terms in the sequence of partial sums, we can see that
S1=cos(1)−cos(12)
S2=(cos(1)−cos(12))+(cos(12)−cos(13))=cos(1)−cos(13)
S3=(cos(1)−cos(12))+(cos(12)−cos(13))+(cos(13)−cos(14))
=cos(1)−cos(14).
In general,
Sk=cos(1)−cos(1k+1).
Since 1/(k+1)→0 as k→∞ and cosx is a continuous function, cos(1/(k+1))→cos(0)=1. Therefore, we conclude that Sk→cos(1)−1. The telescoping series converges and the sum is given by
∞∑n=1[cos(1n)−cos(1n+1)]=cos(1)−1.
Exercise 4.2.5
Determine whether ∞∑n=1[e1/n−e1/(n+1)] converges or diverges. If it converges, find its sum.
- Hint
-
Write out the sequence of partial sums to see which terms cancel.
- Answer
-
e−1
Euler’s Constant
We have shown that the harmonic series ∞∑n=11n diverges. Here we investigate the behavior of the partial sums Sk as k→∞. In particular, we show that they behave like the natural logarithm function by showing that there exists a constant γ such that
\displaystyle \sum_{n=1}^k\left(\frac{1}{n}−\ln k\right)→γ as k→∞.
This constant γ is known as Euler’s constant.
1. Let \displaystyle T_k=\sum_{n=1}^k\left(\frac{1}{n}−\ln k\right). Evaluate T_k for various values of k.
2. For T_k as defined in part 1. show that the sequence {T_k} converges by using the following steps.
a. Show that the sequence {T_k} is monotone decreasing. (Hint: Show that \ln(1+1/k>1/(k+1))
b. Show that the sequence {T_k} is bounded below by zero. (Hint: Express \ln k as a definite integral.)
c. Use the Monotone Convergence Theorem to conclude that the sequence {T_k} converges. The limit γ is Euler’s constant.
3. Now estimate how far T_k is from γ for a given integer k. Prove that for k≥1, 0<T_k−γ≤1/k by using the following steps.
a. Show that \ln(k+1)−\ln k<1/k.
b. Use the result from part a. to show that for any integer k,
T_k−T_{k+1}<\frac{1}{k}−\frac{1}{k+1}. \nonumber
c. For any integers k and j such that j>k, express T_k−T_j as a telescoping sum by writing
T_k−T_j=(T_k−T_{k+1})+(T_{k+1}−T_{k+2})+(T_{k+2}−T_{k+3})+⋯+(T_{j−1}−T_j). \nonumber
Use the result from part b. combined with this telescoping sum to conclude that
T_k−T_j<\frac{1}{k}−\frac{1}{j}. \nonumber
a. Apply the limit to both sides of the inequality in part c. to conclude that
T_k−γ≤\frac{1}{k}. \nonumber
e. Estimate γ to an accuracy of within 0.001.
Key Concepts
- Given the infinite series
\displaystyle \sum_{n=1}^∞a_n=a_1+a_2+a_3+⋯
and the corresponding sequence of partial sums {S_k} where
\displaystyle S_k=\sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k,
the series converges if and only if the sequence {S_k} converges.
- The geometric series \displaystyle \sum^∞_{n=1}ar^{n−1} converges if |r|<1 and diverges if |r|≥1. For |r|<1,
\displaystyle \sum_{n=1}^∞ar^{n−1}=\frac{a}{1−r}.
- The harmonic series
\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+⋯
diverges.
- A series of the form \displaystyle \sum_{n=1}^∞[b_n−b_{n+1}]=[b_1−b_2]+[b_2−b_3]+[b_3−b_4]+⋯+[b_n−b_{n+1}]+⋯ is a telescoping series. The k^{\text{th}} partial sum of this series is given by S_k=b_1−b_{k+1}. The series will converge if and only if \displaystyle \lim_{k→∞} b_{k+1} exists. In that case,
\displaystyle \sum_{n=1}^∞[b_n−b_{n+1}]=b_1−\lim_{k→∞}(b_{k+1}).
Key Equations
- Harmonic series
\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+⋯
- Sum of a geometric series
\displaystyle \sum_{n=1}^∞ar^{n−1}=\frac{a}{1−r} for |r|<1
Glossary
- convergence of a series
- a series converges if the sequence of partial sums for that series converges
- divergence of a series
- a series diverges if the sequence of partial sums for that series diverges
- geometric series
- a geometric series is a series that can be written in the form
\displaystyle \sum_{n=1}^∞ar^{n−1}=a+ar+ar^2+ar^3+⋯
- harmonic series
- the harmonic series takes the form
\displaystyle \sum_{n=1}^∞\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+⋯
- infinite series
- an infinite series is an expression of the form
\displaystyle a_1+a_2+a_3+⋯=\sum_{n=1}^∞a_n
- partial sum
-
the k^{\text{th}} partial sum of the infinite series \displaystyle \sum^∞_{n=1}a_n is the finite sum
\displaystyle S_k=\sum_{n=1}^ka_n=a_1+a_2+a_3+⋯+a_k
- telescoping series
- a telescoping series is one in which most of the terms cancel in each of the partial sums