summation problem

**hatsoff** · May 14th 2008, 03:38 PM

Hi, all. I need to evaluate the following:

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

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!

**Mathstud28** · May 14th 2008, 04:25 PM

Originally Posted by hatsoff

Hi, all. I need to evaluate the following:

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

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

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

Get it now?

**hatsoff** · May 14th 2008, 04:41 PM

Originally Posted by Mathstud28

I will give you a hint

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

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

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.

**Mathstud28** · May 14th 2008, 04:44 PM

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

**hatsoff** · May 14th 2008, 04:52 PM

Originally Posted by Mathstud28

Think derivative/integration

I need another hint, sir.

**Isomorphism** · May 14th 2008, 07:34 PM

Originally Posted by hatsoff

Hi, all. I need to evaluate the following:

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

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

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

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

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

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

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

**Mathstud28** · May 14th 2008, 07:36 PM

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

**hatsoff** · May 14th 2008, 07:47 PM

Originally Posted by Mathstud28

Think derivative/integration

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

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

?

If so, then:

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

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

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

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

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

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

**Mathstud28** · May 14th 2008, 07:50 PM

Originally Posted by hatsoff

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

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

?

If so, then:

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

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

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

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

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

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

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

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

The same applies for derivatives

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

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

**topsquark** · May 14th 2008, 08:00 PM

Originally Posted by Mathstud28

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

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

The same applies for derivatives

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

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

**Mathstud28** · May 14th 2008, 08:02 PM

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

**hatsoff** · May 14th 2008, 08:11 PM

Thanks, guys! This was a big help.

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