Pandora’s Box

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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}} = {{\sum\limits_{i=1}^{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.

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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)

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Fig. 1

At ACA 2017 in Jerusalem, I gave a talk on “Generating Power Summation Formulas using a Computer Algebra System“.

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.

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