6.5: Some classical inequalities

The fact that our formula for the sum of a geometric series gives us an exact sum is very unusual - and hence very precious. For almost all other infinite series - no matter how natural, or beautiful, they may seem - you can be fairly sure that there is no obvious exact formula for the value of the sum. Hence in those cases where we happen to know the exact value, you may infer that it took the best efforts of some of the finest mathematical minds to discover what we know.

One way in which we can make a little progress in estimating the value of an infinite series is to obtain an inequality by comparing the given sum with a geometric series.

Problem 246

(a)(i) Explain why

$\frac{1}{3^2}<\frac{1}{2^2}$,

so$\frac{\mathrm{}}{}$

$\frac{1}{2^2}+\frac{1}{3^2}<\frac{2}{2^2}=\frac{1}{2}$.

(ii) Explain why $\frac{1}{5^2},\frac{1}{6^2},\frac{1}{7^2}$ are all $<\frac{1}{4^2}$, so$\frac{\mathrm{}}{}$

$\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}<\frac{4}{4^2}=\frac{1}{4}.$

(b) Use part (a) to prove that

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}<2,\; for\; all$ $$$n⩾1.

(c) Conclude that the endless sum

$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +$$1n2+⋯ (for ever)

has a definite value, and that this value lies somewhere between $\frac{17}{12}$ and 2.

The next problem presents a rather different way of deriving a similar equality. Once the relevant inequality has been guessed, or given (see Problem 247(a) and (b)), the proof by mathematical induction is often relatively straightforward. And after a little thought about Problem 246, it should be clear that much of the inaccuracy in the general inequality arises from the rather poor approximations made for the first few terms (when n = 1, when n = 2, when n = 3, etc.); hence by keeping the first few terms as they are, and only approximating for $$n⩾2, or $$n⩾3, or $$n⩾4, we can often prove a sharper result.

Problem 247

(a) Prove by induction that$\frac{\mathrm{}}{}$

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{{\mathit{n}}^2}⩽2-\frac{1}{\mathit{n}}$, for all $$n⩾1.

(b) Prove by induction that$\frac{\mathrm{}}{}$

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{}{}$1n2<1.68−1n for all $$n⩾4.

The infinite sum

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{}{}$1n2+⋯ (for ever)

is a historical classic, and has many instructive stories to tell. Recall that, in Problems 54, 62, 63, 236, 237, 238 you found closed formulae for the sums

$1+2+3+\cdots +\mathit{n}$${\mathrm{}}^{}$

$1^2+2^2+3^2+\cdots +{\mathit{n}}^2$${\mathrm{}}^{}$

$1^3+2^3+3^3+\cdots +{\mathit{n}}^3$

and for the sums$\mathrm{}$

$1\times 2+2\times 3+3\times 4+\cdots +(\mathit{n}-1)\mathit{n}$$\mathrm{}$

$1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+\cdots +(\mathit{n}-2)(\mathit{n}-1)\mathit{n}$.

Each of these expressions has a “natural” feel to it, and invites us to believe that there must be an equally natural compact answer representing the sum. In Problem 235 you took this idea one step further by finding a beautiful closed expression for the sum

$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+\cdots +\frac{1}{\mathit{n}(\mathit{n}+1)}=1-\frac{1}{\mathit{n}+1}$

When we began to consider infinite series, we found the elegant closed formula

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots +{\mathit{r}}^{\mathit{n}}=\frac{1}{1-\mathit{r}}-\frac{{\mathit{r}}^{\mathit{n}+1}}{1-\mathit{r}}$.

We then observed that the final term on the RHS could be viewed as an “error term”, indicating the amount by which the LHS differs from $\frac{1}{1-\mathit{r}}$, and noticed that, for any given value of r between −1 and +1, this error term “tends towards 0 as the power n increases”. We interpreted this as indicating that one could assign a value to the endless sum

$1+\mathit{r}+{\mathit{r}}^2+{\mathit{r}}^3+\cdots$ (for ever) = $\frac{1}{1-\mathit{r}}$.

In the same way, in the elegant closed formula

$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+\cdots +\frac{1}{\mathit{n}(\mathit{n}+1)}=1-\frac{1}{\mathit{n}+1}$

the final term on the RHS indicates the amount by which the finite sum on the LHS differs from 1; and since this “error term” tends towards 0 as n increases, we may assign a value to the endless sum

$\frac{1}{1·2}+\frac{1}{2·3}+\frac{1}{3·4}+\cdots$ (for ever) = 1.

It is therefore natural to ask whether other infinite series, such as

$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{}{}$1n2+⋯ (for ever)

may also be assigned some natural finite value. And since the series is purely numerical (without any variable parameters, such as the “r” in the geometric series formula), this answer should be a strictly numerical answer. And it should be exact - though all we have managed to prove so far (in Problems 246 and 247) is that this numerical answer lies somewhere between $\frac{17}{12}$ and 1.68.

This question arose naturally in the middle of the seventeenth century, when mathematicians were beginning to explore all sorts of infinite series (or “sums that go on for ever”). With a little more work in the spirit of Problems 246 and 247 one could find a much more accurate approximate value. But what is wanted is an exact expression, not an unenlightening decimal approximation. This aspiration has a serious mathematical basis, and is not just some purist preference for elegance. The actual decimal value is very close to

$1.649934\cdots$.

But this conveys no structural information. One is left with no hint as to why the sum has this value. In contrast, the eventual form of the exact expression suggests connections whose significance remains of interest to this day.

The greatest minds of the seventeenth and early eighteenth century tried to find an exact value for the infinite sum - and failed. The problem became known as the Basel problem (after Jakob Bernoulli (1654-1705) who popularised the problem in 1689 - one of several members of the Bernoulli family who were all associated with the University in Basel). The problem was finally solved in 1735 - in truly breathtaking style - by the young Leonhard Euler (1707-1783) (who was at the time also in Basel). The answer

$\frac{}{}$π26

illustrates the final sentence of the preceding paragraph in unexpected ways, which we are still trying to understand.

In the next problem you are invited to apply similar ideas to an even more important series. Part (a) provides a relatively crude first analysis. Part (b) attacks the same question; but it does so using algebra and induction (rather than the formula for the sum of a geometric series) in a way that is then further refined in part (c).

Problem 248

(a)(i) Choose a suitable r and prove that$\frac{\mathrm{}}{}$

$\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}<1+\mathit{r}+{\mathit{r}}^2+\cdots +{\mathit{r}}^{\mathit{n}-1}<2$.

(ii) Conclude that

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}<3$ for every $$n⩾0,

and hence that the endless sum

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}+\cdots$ (forever)

can be assigned a value “e” satisfying $2<$e⩽3.

(b) (i) Prove by induction that

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}⩽3-\frac{1}{\mathit{n}.\mathit{n}!}$, for all $$n⩾1.

(ii) Use part (i) to conclude that the endless sum

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}+\cdots$ (forever)

can be assigned a definite value “e”, and that this value lies somewhere between 2.5 and 3.

(c) (It may help to read the Note at the start of the solution to part (c) before attempting parts (c), (d).)

(i) Prove by induction that$\frac{\mathrm{}}{}$

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}⩽2.75-\frac{1}{\mathit{n}.\mathit{n}!}$ for all $$n⩾2.

(ii) Use part (i) to conclude that the endless sum

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}+\cdots$ (for ever)

can be assigned a definite value “e”, and that this value lies somewhere between 2.6 and 2.75.

(d) (i) Prove by induction that

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}⩽2.722\cdots (\mathrm{forever})-\frac{1}{\mathit{n}.\mathit{n}!},$foralln⩾3.

(ii) Use part (i) to conclude that the endless sum

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots +\frac{1}{\mathit{n}!}+\cdots$ (for ever)

can be assigned a definite value “e”, and that this value lies somewhere between 2.708 and 2.7222... (forever).

We end this section with one more inequality in the spirit of this section, and two rather different inequalities whose significance will become clear later.

Problem 249 Prove by induction that

$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +$1n⩾n, forall $$n⩾1.

Problem 250 Let a, b be real numbers such that $\mathit{a}\ne \mathit{b}$, and $\mathit{a}+\mathit{b}>0$. Prove by induction that

$2^{\mathit{n}-1}\left({\mathit{a}}^{\mathit{n}}+\right)$bn⩾(a+b)n, for all $$n⩾1.

Problem 251 Let x be any real number $⩾-1$. Prove by induction that

$(1+\mathit{x}{)}^{}$n⩾1+nx, for all $$n⩾1