Pandora’s Box

Summations arise regularly in mathematical analysis. For example,

$\sum\limits_{i=1}^{n}{1 \over {i (i+1)}} = {n \over {n+1}}$

Having a simple closed form expression such as ${n \over {n+1}}$ makes the summation easier to understand and evaluate.

The summation we focus on in this post is

$\sum\limits_{i=1}^{n}i 2^i\quad\quad\quad(1)$

We will find a closed form for it.

In a recent post, I derived the closed form of a simpler summation (see “Beer theorems and their proofs“) Namely,

$\sum\limits_{i=0}^{n}x^i={{x^{n+1}-1} \over {x-1}}\quad\quad\quad(2)$

From (2) it follows that

${d \over {dx}}{\sum\limits_{i=0}^{n}x^i} = {d \over {dx}}({ {x^{n+1}-1} \over {x-1} })$

which gives us

${\sum\limits_{i=0}^{n}{{d \over dx}x^i}}={{(n+1)x^{n}(x-1)-(x^{n+1}-1)} \over {(x-1)^2}}$ .

Or,

${\sum\limits_{i=0}^{n}{i x^{i-1}}}= {{\sum\limits_{i=0}^{n}{i x^{i}}} \over {x}}$

$= {{(n+1)x^{n}(x-1)-(x^{n+1}-1)} \over {(x-1)^2}}$ .

Therefore,

${\sum\limits_{i=1}^{n}{i x^{i}}}={{(n+1)x^{n+1}(x-1)-x^{n+2}+x} \over {(x-1)^2}}$ .

Let $x=2,$ we arrived at (1)’s closed form:

${\sum\limits_{i=1}^{n}i 2^i} = {{(n+1)2^{n+1} -2 ^{n+2} + 2} \over {2-1}} = 2^{n+1} (n-1) + 2$ .

I have a Computer Algebra aided solution too.

Let $s_n \triangleq \sum\limits_{i=1}^{n} i x^i$ ,

we have $s_1 = x, s_{n}-s_{n-1}=n x^n$

Therefore, the closed form of $s_n$ is the solution of initial-value problem

$\begin{cases} {s_{n}-s_{n-1} }= {n x^n} \\ s_1=x\end{cases}$

It is solved by Omega CAS Explorer (see Fig. 1)

Screen Shot 2018-09-21 at 3.24.33 PM.png

Fig. 1

I had a dream that night. In the dream, I was taking a test.

It reads:

Derive the closed form for

$\sum\limits_{i=1}^{n} {1 \over {(3i-2)(3i+1)}}$

$\sum\limits_{i=1}^{n} {1 \over {(2i+1)^2-1}}$

$\sum\limits_{i=1}^{n} {i \over {(4i^2-1)^2}}$

$\sum\limits_{i=1}^{n} {{i^2 4^i} \over {(i+1)(i+2)}}$

$\sum\limits_{i=1}^{n} { i \cdot i!}$

I woke up with a sweat.

Vroom