9.2: Tests for Convergence

There are many ways to determine if a sequence converges—two are listed below. In all cases changing or removing a finite number of terms in a sequence does not affect its convergence or divergence:

The Comparison Test makes sense intuitively, since something larger than a quantity going to infinity must also go to infinity. The Monotone Bounded Test can be understood by thinking of a bound on a sequence as a wall that the sequence can never pass, as in Figure [fig:mbtest]. The increasing sequence \(\seq{a_n}\) in the figure moves toward \(M\) but can never pass it. The sequence thus cannot diverge to \(\infty\), and it cannot fluctuate back and forth since it always increases. Thus it must converge somewhere before or at \(M\).⁴ Notice that the Monotone Bounded Test tells you only that the sequence converges, not what it converges to.

Show that the sequence \(\seq{a_n}_{n=1}^{\infty}\) defined for \(n\ge 1\) by

\[a_n ~=~ \frac{1 \,\cdot\, 3 \,\cdot\, 5 \,\cdots\, (2n-1)} {2 \,\cdot\, 4 \,\cdot\, 6 \,\cdots\, (2n)} \nonumber \]

is convergent.

Solution: Notice that \(\seq{a_n}\) is always decreasing, since

\[a_{n+1} ~=~ \frac{1 \,\cdot\, 3 \,\cdot\, 5 \,\cdots\, (2n-1) \,\cdot\, (2n+1)} {2 \,\cdot\, 4 \,\cdot\, 6 \,\cdots\, (2n)\,\cdot\, (2n+2)} ~=~ a_n \,\cdot\, \frac{2n+1}{2n+2} ~<~ a_n \,\cdot\, (1) ~=~ a_n \nonumber \]

for \(n\ge 1\). The sequence is also bounded, since \(0 < a_n\) and

\[a_n ~=~ \frac{1}{2} \,\cdot\, \frac{3}{4} \,\cdot\, \frac{5}{6} \,\cdots\, \frac{2n-1}{2n} ~<~ 1 \quad\text{for $n\ge 1$} \nonumber \]

since each fraction in the above product is less than 1. Thus, by the Monotone Bounded Test the sequence is convergent.
Note that for a decreasing sequence only the lower bound is needed for the Monotone Bounded Test, not the upper bound. Similarly, for an increasing sequence only the upper bound matters.

Some tests for convergence of a series are listed below:

Most of the above tests have fairly short proofs or at least intuitive explanations. For example, the n-th Term Test follows from the definition of convergence of a series: if \(\sum a_n\) converges to a number \(L\) then since each term \(a_n = s_n - s_{n-1}\) is the difference of successive partial sums, taking the limit yields

\[\lim_{n \to \infty} \,a_n ~=~ \lim_{n \to \infty} \,(s_n - s_{n-1}) ~=~ L - L ~=~ 0 \nonumber \]

by definition of the convergence of a series. \(~\checkmark\)
Since the n-th Term Test can never be used to prove convergence of a series, it is often stated in the following logically equivalent manner:

Show that \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{n}{2n+1} ~=~ \dfrac{1}{3} + \dfrac{2}{5} + \dfrac{3}{7} + \cdots~\) is divergent.

Solution: Since

\[\lim_{n \to \infty} \;\frac{n}{2n+1} ~=~ \frac{1}{2} ~\ne ~ 0 \nonumber \]

then by the n-th Term Test the series diverges.

The Ratio Test takes a bit more effort to prove.⁵ When the ratio \(R\) in the Ratio Test is larger than 1 then that means the terms in the series do not approach 0, and thus the series diverges by the n-th Term Test. When \(R=1\) the test fails, meaning it is inconclusive—another test would need to be used. When the test shows convergence it does not tell you what the series converges to, merely that it converges.

Determine if \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{n}{2^n}~\) is convergent.

Solution: For the series general term \(a_n = \frac{n}{2^n}\),

\[R ~=~ \lim_{n \to \infty} \;\frac{a_{n+1}}{a_n} ~=~ \lim_{n \to \infty} \;\dfrac{\dfrac{n+1}{2^{n+1}}}{\dfrac{n}{2^n}} ~=~ \lim_{n \to \infty} \;\dfrac{n+1}{2n} ~=~ \frac{1}{2} ~<~ 1 ~, \nonumber \]

so by the Ratio Test the series converges.

Figure [fig:integraltest] shows why the Integral Test works.

In Figure [fig:integraltest](a) the area \(\int_1^{\infty} f(x)\,\dx\) is greater than the total area \(S\) of all the rectangles under the curve. Since each rectangle has height \(a_n\) and width \(1\), then \(S=\sum_2^{\infty} a_n\). Thus, since removing the single term \(a_1\) from the series does not affect the convergence or divergence of the series, the series converges if the improper integral converges, and conversely the integral diverges if the series diverges. Similarly, in Figure [fig:integraltest](b) the area \(\int_1^{\infty} f(x)\,\dx\) is less than the total area \(S=\sum_1^{\infty} a_n\) of all the rectangles, so the integral converges if the series converges, and the series diverges if the integral diverges. Notice how in both graphs the rectangles are either all below the curve or all protrude above the curve due to \(f(x)\) being a decreasing function.

Example \(\PageIndex{1}\): pseries

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Solution

Show that the p-series \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{1}{n^p}~\) converges for \(p>1\) and diverges for \(p=1\).

\[\sum_{n=1}^{\infty} \,\frac{1}{n} ~=~ 1 \;+\; \frac{1}{2} \;+\; \frac{1}{3} \;+\; \frac{1}{4} \;+\; \cdots \nonumber \]

thus diverges even though \(a_n = \frac{1}{n} \rightarrow 0\) (which is hence not a sufficient condition for a series to converge).

Note that this example partly proves the p-series Test. The remaining case (\(p < 1\)) is left as an exercise.

The divergence part of the Comparison Test is clear enough to understand, but for the convergence part with \(0 \le a_n \le b_n\) for all \(n\) larger than some \(N\), ignore the (finite) number of terms before \(a_N\) and \(b_N\). Since \(\sum b_n\) converges then its partial sums must be bounded. The partial sums for \(\sum a_n\) then must also be bounded, since \(0 \le a_n \le b_n\) for \(n > N\). So since \(a_n \ge 0\) means that the partial sums for \(\sum a_n\) are increasing, by the Monotone Bounded Test the partial sums for \(\sum a_n\) must converge, i.e. \(\sum a_n\) is convergent.

Determine if \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{1}{n^n}~\) is convergent.

Solution: Since \(n^n \ge n^2 > 0\) for \(n > 2\), then

\[0 ~<~ \frac{1}{n^n} ~\le~ \frac{1}{n^2} \nonumber \]

for \(n > 2\). Thus, since \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) converges (by the p-series Test with \(p=2>1\), as in Example

Example \(\PageIndex{1}\): pseries

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Solution

), the series \(\sum_{n=1}^{\infty} \frac{1}{n^n}\) converges by the Comparison Test.

For the Limit Comparison Test with \(\frac{a_n}{b_n} \rightarrow L < \infty\) and \(L > 0\), by definition of the limit of a sequence, \(\frac{a_n}{b_n}\) can be made arbitrarily close to \(L\). In particular there is an integer \(N\) such that

\[\frac{L}{2} ~<~ \frac{a_n}{b_n} ~<~ \frac{3L}{2} \nonumber \]

for all \(n > N\). Then

\[0 ~<~ a_n ~<~ \frac{3L}{2}\,b_n \quad\text{and $\quad\sum b_n$ converges} \quad\Rightarrow\quad \text{$\sum a_n$ converges} \nonumber \]

by the Comparison test. Likewise,

\[0 ~<~ \frac{L}{2}\,b_n ~<~ a_n \quad\text{and $\quad\sum b_n$ diverges} \quad\Rightarrow\quad \text{$\sum a_n$ diverges} \nonumber \]

by the Comparison Test again. The cases \(L=0\) and \(L=\infty\) are handled similarly.

Determine if \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{n+3}{n \,\cdot\, 2^n}~\) is convergent.

Solution: Since \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) is convergent (as part of a geometric progression) and

\[\lim_{n \to \infty} ~\frac{(n+3)/(n \,\cdot\, 2^n)}{1/2^n} ~=~ \lim_{n \to \infty} ~\frac{n+3}{n} ~=~ 1 \nonumber \]

then by the Limit Comparison Test \(~\sum_{n=1}^{\infty} \frac{n+3}{n \,\cdot\, 2^n}\) is convergent..

A series \(\sum a_n\) is telescoping if \(a_n = b_n - b_{n+1}\) for some sequence \(\seq{b_n}\). Assume the series \(\sum a_n\) and sequence \(\seq{b_n}\) both start at \(n=1\). Then the partial sum \(s_n\) for \(\sum a_n\) is

\[s_n ~=~ a_1 \;+\; a_2 \;+\; \cdots \;+\; a_n ~=~ (b_1 - b_2) \;+\; (b_2 - b_3) \;+\; \cdots \;+\; (b_n - b_{n+1}) ~=~ b_1 - b_{n+1} \nonumber \]

for \(n \ge 1\). Thus, since \(b_1\) is a fixed number, \(\lim_{n \to \infty} s_n\) exists if and only if \(\lim_{n \to \infty} b_{n+1}\) exists, i.e. \(\sum a_n\) converges if and only if \(\seq{b_n}\) converges. So if \(b_n \rightarrow L\) then \(s_n \rightarrow b - L\), i.e. \(\sum a_n\) converges to \(L\), which proves the Telescoping Series Test. Note that the number \(b_1\), as the first number in the sequence \(\seq{b_n}\), could be replaced by whatever the first number is, in case the index \(n\) starts at a number different from 1.

Example \(\PageIndex{1}\): telescoping

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Solution

Determine if \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{1}{n\,(n+1)}~\) is convergent. If it converges then can you find its sum?

Solution: For the sequence \(\seq{b_n}\) with \(b_n = \frac{1}{n}\) for \(n \ge 1\), each term in the series can be written as

\[\frac{1}{n\,(n+1)} ~=~ \frac{1}{n} ~-~ \frac{1}{n+1} ~=~ b_n \;-\; b_{n+1} \nonumber \]

Thus, since \(\seq{b_n}\) converges to 0, by the Telescoping Series Test the series also converges, to \(b_1 - 0 = 1\).

Convergent series have the following properties (based on similar properties of limits):

[sec9dot2]

For Exercises 1-5 show that the given sequence \(\seq{a_n}_{n=1}^{\infty}\) is convergent.

3

\(a_n = \dfrac{2 \,\cdot\, 4 \,\cdot\, 6 \,\cdots\, (2n)} {3 \,\cdot\, 5 \,\cdot\, 7 \,\cdots\, (2n+1)}\vphantom{\dfrac{2^n}{n!}}\)

\(a_n = 1 \,-\, \dfrac{2^n}{n!}\)

\(a_n = \dfrac{2 \,\cdot\, 4 \,\cdot\, 6 \,\cdots\, (2n)} {1 \,\cdot\, 3 \,\cdot\, 5 \,\cdots\, (2n-1)} \;\cdot\; \dfrac{1}{2n+2}\vphantom{\dfrac{2^n}{n!}}\)

2

\(a_n = \dfrac{1}{n}\,\left(\dfrac{2 \,\cdot\, 4 \,\cdot\, 6 \,\cdots\, (2n)} {1 \,\cdot\, 3 \,\cdot\, 5 \,\cdots\, (2n-1)}\right)^2\)

[exer:wallisinv] \(a_n = \dfrac{1}{2} \,\cdot\, \dfrac{3}{2} \,\cdot\, \dfrac{3}{4} \,\cdot\, \dfrac{5}{4} \,\cdot\, \dfrac{5}{6} \,\cdot\, \dfrac{7}{6} \,\cdots\, \dfrac{2n-1}{2n} \,\cdot\, \dfrac{2n+1}{2n}\vphantom{\left(\dfrac{2n}{2n}\right)^2}\)

For Exercises 6-17 determine whether the given series is convergent. [[1.]]

4

\(\bigsum{n = 0}{\infty}~ \sin\,\left(\dfrac{n \pi}{2}\right)\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{n\,(n + 2)}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{2n}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{n}{(n + 1)\,2^n}\)

4

\(\bigsum{n = 2}{\infty}~ \dfrac{1}{n\,\sqrt{\ln \,n}}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{n\,!}{(2n)\,!}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{n}{e^n}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{\cosh^2 n}\)

4

\(\bigsum{n = 1}{\infty}~ \dfrac{n\,!}{2^n}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{\sqrt{n}}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{n\,(2n-1)}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{\ln\,(n+1)}{n^2}\)

For Exercises 18-21 determine whether the given series is convergent. If convergent then find its sum. [[1.]]

4

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{(2n+1)\,(2n+3)}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{(2n + 3)\,(2n + 5)}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{2}{(3n + 1)\,(3n + 4)}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{1}{4n^2 - 1}\)

[[1.]]

Continue Example

Example \(\PageIndex{1}\): pseries

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Solution

with a proof of the p-series Test for \(p < 1\).

Show that \(\seq{a_n}_{n=1}^{\infty}\) is convergent, where

\[a_n ~=~ \frac{1}{1!} \;+\; \frac{1}{2!} \;+\; \frac{1}{3!} \;+\; \frac{1}{4!} \;+\; \cdots \;+\; \frac{1}{n!} \nonumber \]

for \(n \ge 1\). (Hint: Use the Monotone Bounded test by using a bound on \(\frac{1}{n!}\) for \(n > 2\).)

Consider the series \(~\bigsum{n = 1}{\infty}~ \dfrac{1}{2n-1} ~=~ 1 \;+\; \frac{1}{3} \;+\; \frac{1}{5} \;+\; \frac{1}{7} \;+\; \dotsb ~\).

1. Show that the series is divergent.
2. The textbook Applied Mathematics for Physical Chemistry (3^rd ed.), J. Barrante, provides the following argument that the above series converges: Since

   \[1 ~+~ \frac{1}{4} ~+~ \frac{1}{9} ~+~ \frac{1}{16} ~+~ \dotsb \quad < \quad 1 ~+~ \frac{1}{3} ~+~ \frac{1}{5} ~+~ \frac{1}{7} ~+~ \dotsb \quad < \quad 1 ~+~ \frac{1}{2} ~+~ \frac{1}{3} ~+~ \frac{1}{4} ~+~ \dotsb \nonumber \]

   where the series on the left converges (by the p-series Test with \(p = 2\)) and the series on the right diverges (by the p-series Test with \(p = 1\)), and since each term in the middle series is between its corresponding terms in the left series and right series, then there must be a p-series for some value \(1 < p < 2\) such that each term in the middle series is less than the corresponding term in that p-series. That is,

   \[1 ~+~ \frac{1}{4} ~+~ \frac{1}{9} ~+~ \frac{1}{16} ~+~ \dotsb \quad < \quad 1 ~+~ \frac{1}{3} ~+~ \frac{1}{5} ~+~ \frac{1}{7} ~+~ \dotsb \quad < \quad 1 ~+~ \frac{1}{2^p} ~+~ \frac{1}{3^p} ~+~ \frac{1}{4^p} ~+~ \dotsb \nonumber \]

   for that value of \(p\) between 1 and 2. But \(p > 1\) for that p-series on the right, so it converges, which means that the middle series converges! Find and explain the flaw in this argument.

Wallis’ formula⁶ for \(\pi\) is given by the infinite product

\[\frac{\pi}{2} ~=~ \dfrac{2}{1} \,\cdot\, \dfrac{2}{3} \,\cdot\, \dfrac{4}{3} \,\cdot\, \dfrac{4}{5} \,\cdot\, \dfrac{6}{5} \,\cdot\, \dfrac{6}{7} \,\cdots\, \dfrac{2n}{2n-1} \,\cdot\, \dfrac{2n}{2n+1} \,\cdot\, \cdots ~. \nonumber \]

Notice that this is the limit of the reciprocal of the sequence in Exercise [exer:wallisinv]. Write a computer program to approximate the limit using 1 million iterations. How close is your approximation to \(\frac{\pi}{2}\)?