# Thread: prove that these series converge to values given

1. ## prove that these series converge to values given

I know how to determine what the series $\sum_{n=1}^{\infty}n^pr^{n-1}$ converge to (for -1 < r < 1) for positive integral values of p:

p = 1: $\frac{1}{(1 - r)^2}$

p = 2: $\frac{1+r}{(1 - r)^3}$

p = 3: $\frac{1 + 4r + r^2}{(1 - r)^4}$

p = 4: $\frac{1 + 11r + 11r^2 + r^3}{(1 - r)^5}$

p = 5: $\frac{1 + 26r + 66r^2 + 26r^3 + r^4}{(1 - r)^6}$

The general pattern is:

$\frac{a_0 + a_1r + ... + a_{p-2}r^{p-2} + a_{p-1}r^{p-1}}{(1 - r)^{p+1}}$

where the coefficients $a_0, a_1, ... a_{p-1}$ are given by the pth row of Euler's triangle (there is an article on the triangle at www.mathworld.com).

So I figured this out but I don't know how to prove it formally. Any suggestions?

2. Originally Posted by icemanfan
I know how to determine what the series $\sum_{n=1}^{\infty}n^pr^{n-1}$ converge to for positive integral values of p:

p = 1: $\frac{1}{(1 - r)^2}$

p = 2: $\frac{1+r}{(1 - r)^3}$

p = 3: $\frac{1 + 4r + r^2}{(1 - r)^4}$

p = 4: $\frac{1 + 11r + 11r^2 + r^3}{(1 - r)^5}$

p = 5: $\frac{1 + 26r + 66r^2 + 26r^3 + r^4}{(1 - r)^6}$

The general pattern is:

$\frac{a_0 + a_1r + ... + a_{p-2}r^{p-2} + a_{p-1}r^{p-1}}{(1 - r)^{p+1}}$

where the coefficients $a_0, a_1, ... a_{p-1}$ are given by the pth row of Euler's triangle (there is an article on the triangle at www.mathworld.com).

So I figured this out but I don't know how to prove it formally. Any suggestions?
Here is an idea

First re-index the series to start at zero. this gives

$\displaystyle \sum_{n=0}^{\infty}(n+1)^pr^n$

Now using the binomial theorem of the factor

$\displaystyle (n+1)^p=\sum_{k=0}^{p}\binom{p}{k}n^{k}$

Combing the two above gives

$\displaystyle \sum_{n=0}^{\infty}\sum_{k=0}^{p}\binom{p}{k}n^{k} r^n$

Now lets analyze the pattern of

$\displaystyle k=0; \quad r^n$
$\displaystyle k=1; \quad n^{1}r^n=r \frac{d}{dr}r^n$
$\displaystyle k=2; \quad n^{2}r^n=r \frac{d}{dr}r \frac{d}{dr}r^n$
$\displaystyle k=p; \quad n^pr^n =\left( r\frac{d}{dr}\right)^pr^n$

$\displaystyle \sum_{n=0}^{\infty}\sum_{k=0}^{p}\binom{p}{k}n^{k} r^n=\sum_{n=0}^{\infty}\sum_{k=0}^{p}\binom{p}{k}\ left( r\frac{d}{dr}\right)^kr^n=\sum_{k=0}^{p}\binom{p}{ k}\left( r\frac{d}{dr}\right)^k\sum_{n=0}^{\infty}r^n=\sum_ {k=0}^{p}\binom{p}{k}\left( r\frac{d}{dr}\right)^k\frac{1}{1-r}$

Note that the derivatives can be pulled out of the infinite sum for $|r|<1$ because the geometric series is uniformly convergent.