\( \def\List{\operatorname{List}} \) \( \def\integer{\operatorname{integer}} \) \( \def\nat{\operatorname{nat}} \) \( \def\Bool{\operatorname{Bool}} \) \( \def\Punct{\operatorname{Punct}} \) \( \def\BT{\operatorname{BT}} \) \( \def\B{\operatorname{B}} \) \( \def\T{\operatorname{T}} \) \( \def\R{\operatorname{R}} \) \( \def\Set{\operatorname{Set}} \) \( \def\Bag{\operatorname{Bag}} \) \( \def\Ring{\operatorname{Ring}} \) \( \def\plug{\operatorname{plug}} \) \( \def\Zipper{\operatorname{Zipper}} \) \( \def\cosh{\operatorname{cosh}} \) \( \def\sinh{\operatorname{sinh}} \) \( \def\ch{\operatorname{ch}} \) \( \def\sh{\operatorname{sh}} \) \( \def\d{\partial} \)
a trivial exercise.

Let's find a sum

$$\sum_{n=1}^\infty {1\over{n(n+2)}}$$

There is a well-know standard way, that I managed to recall eventually. Given that

$${1 \over n(n+2)} = {1 \over 2}\cdot \left({1\over n} - {1\over n+2}\right)$$

the sum can be re-written as

$$\sum_{n=1}^\infty {1\over{n(n+2)}} = \sum_{n=1}^\infty {1 \over 2}\left({1\over n} - {1\over n+2}\right) = {1\over 2}\left({1\over 1} - {1\over 3} + {1\over 2} - {1\over 4} + {1\over 3} - {1\over 5} + {1\over 4} - {1\over 6} \cdots\right)$$

with almost all terms canceling each other, leaving

$$\sum_{n=1}^\infty {1\over{n(n+2)}} = {1\over 2}\left(1 + {1\over 2}\right) = {3\over 4}$$

While this is easy to check, very little help is given on understanding how to arrive to the solution in the first place. Indeed, the first (and crucial) step is a rabbit pulled sans motif out of a conjurer hat. The solution, fortunately, can be found in a more systematic fashion, by a relatively generic method. Enter generating functions.

First, introduce a function

$$f(t) = \sum_{n=1}^\infty {t^{n + 1}\over n}$$

The series on the right converge absolutely when \(|t| < 1\), so one can define

$$g(t) = \int f(t) dt = \int \sum_{n=1}^\infty {t^{n + 1}\over n} = \sum_{n=1}^\infty \int {t^{n + 1}\over n} = \sum_{n=1}^\infty {t^{n + 2}\over {n(n+2)}} + C$$

with the sum in question being

$$\sum_{n=1}^\infty {1\over{n(n+2)}} = g(1) - C = g(1) - g(0)$$

Definition of the \(g\) function follows immediately from the form of the original sum, and there is a limited set of operations (integration, differentiation, etc.) applicable to \(g\) to produce \(f\).

The rest is more or less automatic. Note that

$$- ln(1 - t) = t + {t^2\over 2} + {t^3\over 3} + \cdots$$

so that

$$f(t) = t^2 + {t^3\over 2} + {t^4\over 3} + \cdots = - t \cdot ln(1-t)$$

therefore

$$g(t) = - \int t \cdot ln(1-t) dt = \cdots = {1\over 4} (1 - t)^2 - {1\over 2} (1 - t)^2 ln(1 - t) + (1 - t) ln(1 - t) + t + C$$

where the integral is standard. Now,

$$g(1) - g(0) = 1 - {1\over 4} = {3\over 4}$$

Voilà!

And just to check that things are not too far askew, a sub-exercise in pointless programming:

scala> (1 to 10000).map(x => 1.0/(x*(x+2))).reduceLeft(_+_)
res0: Double = 0.749900014997506

PS: of course this post is an exercise in tex2img usage.

PPS: Ed. 2022: tex2img is gone, switch to mathjax.