1. Evaluating this infinite sum

Is it possible to evaluate this infinite sum:

$\sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$

2. Originally Posted by acevipa
Is it possible to evaluate this infinite sum:

$\sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
This way seems messy, but it works..

Skip proving convergence.... it converges.

$\displaystyle L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}\left(\frac{1}{3}\right)$

$\displaystyle 3L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^{i-1}$

$\displaystyle 2L = \sum_{i=1}^\infty i\left(\frac{2}{3}\right)^i$

This form is "cleaner." Write some terms

$2L = \dfrac{2}{3} + 2\left(\dfrac{4}{9}\right) + 3\left(\dfrac{8}{27}\right) + 4\left(\dfrac{16}{81}\right) + \dots$

Consider what happens when we multiply by 3,

$6L = 2 + \dfrac{2\cdot4}{3} + \dfrac{3\cdot8}{9} + \dfrac{4\cdot16}{27} + \dots$

Add the two together, matching up terms with common denominators.

$8L = 2 + \dfrac{2+2\cdot4}{3} + \dfrac{2\cdot4+3\cdot8}{9} + \dfrac{3\cdot8+4\cdot16}{27} + \dots$

$= 2 + 5\left(\dfrac{2}{3}\right) + 8\left(\dfrac{2}{3}\right)^2 + 11\left(\dfrac{2}{3}\right)^3 + \dots$

So

$\displaystyle 8L = 2 + \sum_{i=1}^{\infty}(3i+2)\left(\frac{2}{3}\right)^ i$

$\displaystyle = 2 + \sum_{i=1}^{\infty}3i\left(\frac{2}{3}\right)^i + 2\sum_{i=1}^{\infty}\left(\frac{2}{3}\right)^i$

Simplifying,

$8L = 2 + 3(2L) + 4$

$2L = 6$

$L = 3$

3. For $|x|<1$ is...

$\displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)

... so that is...

$\displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)

Now from (2) follows...

$\displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
For $|x|<1$ is...

$\displaystyle f(x)= \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ (1)

... so that is...

$\displaystyle f^{'} (x) = \sum_{n=1}^{\infty} n\ x^{n-1} = \frac{1}{(1-x)^{2}}$ (2)

Now from (2) follows...

$\displaystyle \sum_{n=1}^{\infty} n\ (\frac{2}{3})^{n-1}\ \frac{1}{3} = \frac{1}{3}\ f^{'} (\frac{2}{3}) = 3$ (3)

Kind regards

$\chi$ $\sigma$
Great method. Thanks

5. Originally Posted by acevipa
Great method. Thanks
If you can follow chisigma's solution you have posted this in the wrong forum and wasted the time of those helpers who have given or may have considered giving solutions that do not use calculus.

CB

6. Originally Posted by acevipa
Is it possible to evaluate this infinite sum:

$\sum_{n=1}^{\infty} n\left(\dfrac{2}{3}\right)^{n-1}\left(\dfrac{1}{3}\right)$
$\displaystyle\ S_{\infty}=\frac{1}{3}+\left(\frac{2}{3}\right)\le ft(\frac{2}{3}\right)+\left(\frac{3}{3}\right)\lef t(\frac{2}{3}\right)^2+\left(\frac{4}{3}\right)\le ft(\frac{2}{3}\right)^3+.....$

$3\displaystyle\ S_{\infty}-1=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+4\left(\frac{2}{3}\right)^3+....$

$s=2\left(\frac{2}{3}\right)+3\left(\frac{2}{3}\rig ht)^2+.....$

$\frac{2}{3}s=2\left(\frac{2}{3}\right)^2+3\left(\f rac{2}{3}\right)^3+4\left(\frac{2}{3}\right)^4....$

$s-\frac{2}{3}s=\frac{s}{3}=2\left(\frac{2}{3}\right) +\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\righ t)^3+.....$

$\frac{s-2}{3}=\left(\frac{2}{3}\right)+\left(\frac{2}{3}\r ight)^2+\left(\frac{2}{3}\right)^3+....$

which is purely a geometric series.
Then use the result of this to calculate the original sum.

i will use

$\Delta f(x)=f(x+1)-f(x)$ for the difference operator

and $\sum\limits_x f(x) = g(x)$ if and only if $\Delta g(x) =f(x)$ (the indefinite summation)

we have the summation by parts method

$\sum\limits_{x} g(x)\Delta
f(x)=f(x).g(x)-\sum\limits_{x} f(x+1).\Delta g(x)$

(its similar to the integration by parts)

if we apply this method "n" times we get
$$\sum\limits_{x} g(x).\Delta f(x)=$$

$\sum\limits^{n}_{k=0}\Delta^{k}g(x).(-1)^{k}\Delta^{-k}E^{k}f(x)+(-1)^{n+1}\sum\limits_{x}\Delta^{n+1}g(x)\Delta^{-n}E^{n+1}f(x)$

( where $E^k f(x)=f(x+k)$)

if, g(x) has degree n , (so $\Delta ^{n+1} g(x)=0$ )and $|a|<1$ we can use it on the summation by parts to find

$\sum\limits^{\infty}_{x=c} g(x)a^{x}=a^{c-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}g(x)|_{x=c} }{(b-1)^{k+1}}$

where $b= \frac{1}{a}$

if $g(x)=x^{n}$ and e c=0
$\sum\limits^{\infty}_{x=0} x^{n}a^{x}=a^{-1}\sum\limits^{n}_{k=0}\frac{\Delta^{k}x^{n}|_{x=0 }}{(b-1)^{k+1}}$

but $\Delta^{k}x^{n}|_{x=0}=k!\bigg\{^{n}_{k}\bigg\}$ (here $\bigg\{^{n}_{k}\bigg\}$ it's the stirling number of the second kind, knuth's notation )
so
$\sum\limits^{\infty}_{x=0} x^{n}a^{x}= a^{-1}\sum\limits^{n}_{k=0}\frac{k! \bigg\{^{n}_{k}\bigg\}}{(b-1)^{k+1}}.$

8. (other way , the general case)

we can define the operator xD, (xD) f(x) =x [D f(x)] ( where "D" is the derivative )
then we can show

$(xD)^{n}g(x)=\sum\limits^{n}_{k=0}\bigg\{^{n}_{k}\ bigg\}x^{k}D^{k}g(x)$

(again the stirling numbers of the second kind)
take the series
$\sum\limits^{\infty}_{n=0}x^{n}=\frac{1}{1-x}$

apply $(xD)^p$ in the both sides

$\sum\limits^{\infty}_{n=0}n^{p}x^{n}=(xD)^p\frac{1 }{1-x}$

using the identity
$(xD)^{p}g(x)=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\ bigg\}x^{k}D^{k}g(x)$

we have

$(xD)^{p}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}x^{k }D^{k}\frac{1}{1-x}=\sum\limits^{p}_{k=0}\bigg\{^{p}_{k}\bigg\}\fra c{k!x^{k}}{(1-x)^{k+1}}$

so
$\sum\limits^{\infty}_{n=0}n^{p}x^{n}=\sum\limits^{ p}_{k=0}\bigg\{^{p}_{k}\bigg\}\frac{k!x^{k}}{(1-x)^{k+1}}.$