If you want to calculate an infinite sum exactly, relate it to coefficients associated with some function

Quick description

In general, one does not expect to be able to calculate an infinite sum $\sum_{n=1}^\infty a_n$ exactly. If one can, there is usually a good reason for it being possible, and very often the reason is that there is a function $f$ that can be described independently of the sequence $(a_n)$ that has coefficients of some kind that are closely related to $(a_n)$ . For example, $f$ might have a power series $f(x)=\sum_{n=1}^\infty c_nx^n$ and a value $\alpha$ such that $c_n\alpha^n=a_n$ for every $n$ (so that the original sum is equal to $f(\alpha)$ ). Or $f$ might have a Fourier expansion $f(x)=\sum_nc_ne^{2\pi ix}$ with $|c_n|^2=a_n$ .

Example 1

An easy example of an infinite series that can be calculated exactly is $\sum_{n=0}^\infty t^n$ , when $|t|<1$ . The usual approach is to calculate explicity the partial sum $\sum_{n=1}^Nt^n$ , using the formula for summing a geometric progression. One gets $(1-t^{N+1})/(1-t)$ , which is easily seen to converge to $1/(1-t)$ .

Note that what we have really done is to solve a whole class of problems, one for each $t\in(-1,1)$ , by proving that the function $1/(1-t)$ has $1+t+t^2+\dots$ as a power-series expansion.

Example 2

Let's see if we can work out the infinite sum $\sum_{n=0}^\infty 2^{-3n}\binom{2n}n$ . The existence of a binomial coefficient might lead us to suspect that if this could be done then the binomial theorem would be involved. This certainly doesn't have to be the case, but we could at least think about ratios of successive terms. The rough size of $\binom{2n}n$ is $n^{-1/2}2^{2n}$ , up to a constant multiple, which increases by a factor of roughly $4$ when $n$ is large and increases by $1$ . Therefore, the terms themselves go down by a factor of roughly $2$ each time.

Now let us compare that with what we know about the binomial expansion of $(1+x)^\alpha$ . We get

$1+\alpha x+\frac{\alpha(\alpha-1)}2x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\dots$

The ratio of the coefficient of $x^n$ to that of $x^{n-1}$ is $(\alpha-n+1)/n$ , which tends to $-1$ , from which we deduce that to get a ratio that tends to $1/2$ we will need $x$ to equal $-1/2$ .

This tells us that the coefficient of $x^n$ will be

$(-1)^n2^{-2n}\binom{2n}n=(-1)^n2^{-2n}\frac{2n!}{(n!)^2}=(-1)^n\frac{1\cdot 3\cdots (2n-1)}{2^nn!}=\frac 1{n!}(-\frac 12)(-\frac 32)\cdots(-\frac{(2n-1)}2).$

Therefore, $\alpha=-1/2$ , so the sum we have calculated is $(1-1/2)^{-1/2}=\sqrt{2}$ .

Looking at this backwards, we can present the following "unjustified" solution to the problem, which shows how the example was thought of.

$(1-x)^{-1/2}=1+\frac{x}2+\frac{1\cdot 3x^2}{2^22!}+\frac{1\cdot 3\cdot 5x^3}{2^33!}+\dots$

so when $x=1/2$ we get

$\begin{align}1+\frac 1{2^2}+\frac {1\cdot 3}{2^42!}+\frac{1\cdot 3\cdot 5}{2^63!}+\dots&=1+\frac{2!}{2^3}+\frac{4!}{2^6(2!)^2}+\frac{6!}{2^9(3!)^2}+\dots\\ &=1+2^{-3}\binom 21+2^{-6}\binom 42+2^{-9}\binom 63+\dots\\ \end{align}$

Example 3

A famous example of a series that can be summed but only with some ingenuity is

$1+\frac 1{2^2}+\frac 1{3^2}+\frac 1{4^2}+\dots = \frac{\pi^2}6,$

a result obtained by Euler. One way of proving it is to define a function $f$ on the interval $[-\pi,\pi]$ by taking $f(x)=-1$ if $x<0$ and $f(x)=1$ if $x\geq 0$ .

What has this to do with the sequence $1,\frac 1{2^2},\frac 1{3^2},\dots$ ? To answer this, let us work out the Fourier coefficients of $f$ . The $n\mathrm{th}$ Fourier coefficient $\hat{f}(n)$ is defined to be $\frac 1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx$ . Now $\int_0^\pi e^{-inx}dx=-\frac 1{in}[e^{-inx}]_0^\pi=-(e^{-in\pi}-1)/in$ . If $n$ is even, then $e^{-in\pi}=1$ , so we get 0. If $n$ is odd, then $e^{-in\pi}=-1$ , and then we get $2/in$ . After a very similar calculation for the part of the integral between $-\pi$ and $0$ , and after taking account of the factor $\frac 1{2\pi}$ we find that $\hat{f}(n)=2/in\pi$ if $n$ is odd and $0$ if $n$ is even.

Now Parseval's identity tells us that the square of the $L_2$ norm of the function $f$ is equal to the sum of the squares of the absolute values of the Fourier coefficients of $f$ . Given the way we have defined the Fourier coefficients, the appropriate definition for the square of the $L_2$ norm of $f$ is $\frac 1{2\pi}\int_{-\pi}^\pi|f(x)|^2dx=1$ .

But the sum of the absolute values of the Fourier coefficients is $\sum_{m=-\infty}^\infty\frac{4}{\pi^2(2m+1)^2}$ . Since the square of an odd number is equal to the square of minus that odd number, we can write this as $\sum_{m=1}^\infty\frac 8{\pi^2(2m-1)^2}$ . Therefore, by Parseval's identity, $\sum_{m=1}^\infty\frac 1{(2m-1)^2}=\frac{\pi^2}8$ .

This looks similar to Euler's theorem. To obtain Euler's theorem itself, we just note that the left-hand side is equal to $\sum_{n=1}^\infty\frac 1{n^2}-\sum_{n=1}^\infty\frac 1{(2n)^2}$ , which is three quarters of $\sum_{n=1}^\infty\frac 1{n^2}$ . So $\sum_{n=1}^\infty\frac 1{n^2}=\frac 43\frac{\pi^2}8$ , which equals $\frac{\pi^2}6$ .

General discussion

In the first two examples above, the coefficients were Taylor coefficients associated with the functions $(1-x)^{-1}$ and $(1-x)^{-1/2}$ , respectively, and we obtained the result by means of evaluation at a point. In the third example, we used Fourier coefficients associated with a step function and obtained the result with the help of Parseval's identity.

Example 4

In the article To calculate an infinite sum exactly, try antidifferencing, the example

$\frac 1{1\cdot 2}+\frac 1{2\cdot 3}+\frac 1{3\cdot 4}+\dots$

was considered, and the usual proof was given that turns it into a telescoping sum. But the question of why this series should be amenable to that treatment was not fully answered. Here is a possible explanation. Let us start with the more general series

$1+x+x^2+x^3+\dots$

which is of a type that we like (because we like geometric progressions). Now we "integrate" twice, obtaining first

$x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\dots$

and then

$\frac {x^2}{1\cdot 2}+\frac{x^3}{2\cdot 3}+\frac{x^4}{3\cdot 4}+\frac{x^5}{4\cdot 5}+\dots$

Forgetting for the moment about convergence questions, let us think what should have happened to the corresponding functions. The original function was $1/(1-x)$ (at least when $|x|<1$ ). Therefore, the next one should be $-\log(1-x)$ , and the last one (after a small calculation) should be $x-(1-x)\log(1-x)$ .

One can check that that all works if $|x|<1$ , that the limit of the last expression as $x$ tends to $1$ is $1$ , and that the limit of the thrid power series above as $x$ tends to $1$ is the sum $\frac 1{1\cdot 2}+\frac 1{2\cdot 3}+\dots$ that we wanted to calculate.

Thus, one explanation for the infinite sum being "nice" is that it is derived from the simplest genuinely infinite sum of all (a geometric progression) by a very natural process (antidifferentiation). But to see this, we again had to look at functions rather than series.