Show that sum x/2^x = 2 summation running 0 to infinity ?

Jul 25, 2017

${\sum}_{n = 0}^{\setminus \infty} \frac{n}{2} ^ n = 2$.

Explanation:

Suppose $\left\{{f}_{n}\right\}$ and $\left\{{g}_{n}\right\}$ are two sequences. Then,

${\sum}_{n = 0}^{N} {f}_{n} \left({g}_{n + 1} - {g}_{n}\right) = {f}_{N + 1} {g}_{N + 1} - {f}_{0} {g}_{0} - {\sum}_{n = 0}^{N} {g}_{n + 1} \left({f}_{n + 1} - {f}_{n}\right)$.

This can be proved by investigating
${\sum}_{n = 0}^{N} {f}_{n} \left({g}_{n + 1} - {g}_{n}\right) + {g}_{n + 1} \left({f}_{n + 1} - {f}_{n}\right)$.

Expanding gives,

${\sum}_{n = 0}^{N} {f}_{n} {g}_{n + 1} - {f}_{n} {g}_{n} + {g}_{n + 1} {f}_{n + 1} - {g}_{n + 1} {f}_{n}$,
${\sum}_{n = 0}^{N} {f}_{n + 1} {g}_{n + 1} - {f}_{n} {g}_{n}$.

This is now a telescoping where all the terms cancel apart from the first and the last, giving

${\sum}_{n = 0}^{N} {f}_{n + 1} {g}_{n + 1} - {f}_{n} {g}_{n} = {f}_{N + 1} {g}_{N + 1} - {f}_{0} {g}_{0}$.

Substituting,

${\sum}_{n = 0}^{N} {f}_{n} \left({g}_{n + 1} - {g}_{n}\right) + {g}_{n + 1} \left({f}_{n + 1} - {f}_{n}\right) = {f}_{N + 1} {g}_{N + 1} - {f}_{0} {g}_{0}$,
${\sum}_{n = 0}^{N} {f}_{n} \left({g}_{n + 1} - {g}_{n}\right) = {f}_{N + 1} {g}_{N + 1} - {f}_{0} {g}_{0} - {\sum}_{n = 0}^{N} {g}_{n + 1} \left({f}_{n + 1} - {f}_{n}\right)$ as required.

Then, let ${f}_{n} = n$ and let ${g}_{n + 1} - {g}_{n} = {2}^{- n}$. We see that if ${g}_{n} = k {2}^{- n}$ then

${g}_{n + 1} - {g}_{n} = k \left({2}^{- \left(n + 1\right)} - {2}^{- n}\right)$,
${g}_{n + 1} - {g}_{n} = k \left(\frac{1}{2} \cdot {2}^{- n} - {2}^{- n}\right)$,
${g}_{n + 1} - {g}_{n} = - \frac{k}{2} {2}^{- n}$.

Then let $k = - 2$ giving ${g}_{n + 1} - {g}_{n} = {2}^{- n}$ for ${g}_{n} = - 2 \cdot {2}^{- n}$

We conclude ${f}_{n} = n$, $g \left(n\right) = - 2 \cdot {2}^{- n}$.

Substituting,

${\sum}_{n = 0}^{N} \frac{n}{2} ^ n = - 2 \left(N + 1\right) {2}^{- \left(N + 1\right)} + 2 {\sum}_{n = 0}^{N} {2}^{- \left(n + 1\right)} \left(n + 1 - n\right)$,
${\sum}_{n = 0}^{N} \frac{n}{2} ^ n = - 2 \left(N + 1\right) {2}^{- \left(N + 1\right)} + {\sum}_{n = 0}^{N} {2}^{- n}$,

By the sum of a geometric series,

${\sum}_{n = 0}^{N} \frac{n}{2} ^ n = - 2 \left(N + 1\right) {2}^{- \left(N + 1\right)} + \frac{1 - {2}^{- \left(N + 1\right)}}{1 - \frac{1}{2}}$,
${\sum}_{n = 0}^{N} \frac{n}{2} ^ n = 2 - 2 \cdot {2}^{- \left(N + 1\right)} \left(\left(N + 1\right) + 1\right)$,
${\sum}_{n = 0}^{N} \frac{n}{2} ^ n = 2 - \left(N + 2\right) {2}^{- N}$

As $N \to + \setminus \infty$ $\left(N + 2\right) {2}^{- N} \to 0$.

Then

${\sum}_{n = 0}^{\setminus \infty} \frac{n}{2} ^ n = 2$.

Jul 25, 2017

See below.

Explanation:

Defining ${S}_{n} = {\sum}_{k = 0}^{n} {x}^{k}$ or

${S}_{n} = \frac{{x}^{n + 1} - 1}{x - 1}$

for $\left\mid x \right\mid < 1$ we have

${S}_{\infty} = {\lim}_{n \to \infty} {S}_{n} = - \frac{1}{x - 1}$

now

${\sum}_{k = 0}^{n} k {x}^{k} = x \left(\frac{{\mathrm{dS}}_{n}}{\mathrm{dx}}\right)$ and

${\sum}_{k = 0}^{\infty} k {x}^{k} = x \left(\frac{{\mathrm{dS}}_{\infty}}{\mathrm{dx}}\right) = x \left(\frac{d}{\mathrm{dx}}\right) \left(\frac{1}{1 - x}\right) = \frac{x}{x - 1} ^ 2$

and now making $x = \frac{1}{2}$ we obtain

${\sum}_{k = 0}^{\infty} k {x}^{k} = 2$