# summation problem

• May 14th 2008, 03:38 PM
hatsoff
summation problem
Hi, all. I need to evaluate the following:

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

where k∈ℕ and k≠1

I don't really know how to do much else than test whether or not a series converges or diverges; I don't know how to determine what it converges to.

Thanks, guys!
• May 14th 2008, 04:25 PM
Mathstud28
Quote:

Originally Posted by hatsoff
Hi, all. I need to evaluate the following:

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

where k∈ℕ and k≠1

I don't really know how to do much else than test whether or not a series converges or diverges; I don't know how to determine what it converges to.

Thanks, guys!

I will give you a hint

Rewrite $\displaystyle \frac{k-1}{k}$ as $\displaystyle 1-\frac{1}{k}$

Now consider that since $\displaystyle k\in\mathbb{N}$ that $\displaystyle 1-\frac{1}{k}<1$

Get it now?
• May 14th 2008, 04:41 PM
hatsoff
Quote:

Originally Posted by Mathstud28
I will give you a hint

Rewrite $\displaystyle \frac{k-1}{k}$ as $\displaystyle 1-\frac{1}{k}$

Now consider that since $\displaystyle k\in\mathbb{N}$ that $\displaystyle 1-\frac{1}{k}<1$

Get it now?

It resembles a geometric series in that respect, but then each term is multiplied by n, which is not a constant. I don't know how to handle that.
• May 14th 2008, 04:44 PM
Mathstud28
Quote:

Originally Posted by hatsoff
It resembles a geometric series in that respect, but then each term is multiplied by n, which is not a constant. I don't know how to handle that.

Think derivative/integration
• May 14th 2008, 04:52 PM
hatsoff
Quote:

Originally Posted by Mathstud28
Think derivative/integration

I need another hint, sir.
• May 14th 2008, 07:34 PM
Isomorphism
Quote:

Originally Posted by hatsoff
Hi, all. I need to evaluate the following:

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

where k∈ℕ and k≠1

I don't really know how to do much else than test whether or not a series converges or diverges; I don't know how to determine what it converges to.

Thanks, guys!

If $\displaystyle |x| < 1$,
$\displaystyle G(x) = \sum_{n=0}^{\infty} x^n = \frac1{1-x}$

$\displaystyle G'(x) = \sum_{n=0}^{\infty} nx^{n-1} = \frac1{(1-x)^2}$

$\displaystyle xG'(x) = \sum_{n=0}^{\infty} nx^{n} = \frac{x}{(1-x)^2}$

Now let $\displaystyle x = 1 - \frac1{k}$, since $\displaystyle k \in \mathbb{N}$, we have $\displaystyle \frac1{k} \leq 1$ and $\displaystyle 0 < 1 - \frac1{k} < 1$. So we can apply the infinte Geometric Progression formula :)

$\displaystyle \sum_{n=0}^{\infty} n(1 - \frac1{k})^{n} = \frac{1 - \frac1{k}}{(\frac1{k})^2}$

$\displaystyle \frac1{k} \sum_{n=0}^{\infty} n(1 - \frac1{k})^{n} = \frac{1 - \frac1{k}}{\frac1{k}} = k - 1$
• May 14th 2008, 07:36 PM
Mathstud28
Quote:

Originally Posted by hatsoff
I need another hint, sir.

Oh I am sorry! I completely missed your response. Luckily for you The oh so competent Isomorphism replied (Giggle)
• May 14th 2008, 07:47 PM
hatsoff
Quote:

Originally Posted by Mathstud28
Think derivative/integration

Hmm. Are you telling me that the following is true:

$\displaystyle \int[\sum_{n=r}^{\infty}a_n]dx=\sum_{n=r}^{\infty}\int[a_ndx]$

?

If so, then:

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

$\displaystyle \frac{1}{k}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^nn$

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

$\displaystyle (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}\int[(\frac{k-1}{k})^{n-1}ndn]$

$\displaystyle (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^{n}$

EDIT: I guess I'm late to the party.
• May 14th 2008, 07:50 PM
Mathstud28
Quote:

Originally Posted by hatsoff
Hmm. Are you telling me that the following is true:

$\displaystyle \int[\sum_{n=r}^{\infty}a_n]dx=\sum_{n=r}^{\infty}\int[a_ndx]$

?

If so, then:

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

$\displaystyle \frac{1}{k}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^nn$

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

$\displaystyle (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}\int[(\frac{k-1} {k})^{n-1}ndn]$

$\displaystyle (\frac{k-1}{k^2})\frac{d}{dn}\sum_{n = 1}^{\infty}(\frac{k-1}{k})^{n}$

EDIT: I guess I'm late to the party.

Yes you are...but for your sake yes..In a power series

$\displaystyle \int\sum_{n=0}^{\infty}a_n(x-c){n}dx=\sum_{n=0}^{\infty}\int{a_n(x-c)^{n}}dx$

The same applies for derivatives

$\displaystyle \frac{D[\sum_{n=0}^{\infty}a_n(x-c)^{n}]}{dx}=\sum_{n=0}^{\infty}\frac{D[a_n(x-c)^{n}]}{dx}$

to prove it expand...differentiate/integrate....contract into an explicit form power series and compare
• May 14th 2008, 08:00 PM
topsquark
Quote:

Originally Posted by Mathstud28
Yes you are...but for your sake yes..In a power series

$\displaystyle \int\sum_{n=0}^{\infty}a_n(x-c){n}dx=\sum_{n=0}^{\infty}\int{a_n(x-c)^{n}}dx$

The same applies for derivatives

$\displaystyle \frac{D[\sum_{n=0}^{\infty}a_n(x-c)^{n}]}{dx}=\sum_{n=0}^{\infty}\frac{D[a_n(x-c)^{n}]}{dx}$

to prove it expand...differentiate/integrate....contract into an explicit form power series and compare

The following phrase should be added to this: "Provided the relevant series converge." If they don't then these rules go out the window.

-Dan
• May 14th 2008, 08:02 PM
Mathstud28
Quote:

Originally Posted by topsquark
The following phrase should be added to this: "Provided the relevant series converge." If they don't then these rules go out the window.

-Dan

Yes! Thank you! Sometimes I have the classic "I already know it so I assume others do to" syndrome...not a good thing to have (Crying)
• May 14th 2008, 08:11 PM
hatsoff
Thanks, guys! This was a big help.