Here is a fun one

Find

$\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n\quad\forall~|x|<1$

Where $\displaystyle a$ is an arbitrary constant.

Results 1 to 14 of 14

- Jun 28th 2008, 10:03 AM #1

- Jun 28th 2008, 11:06 AM #2
$\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}(n^3+1)x^n+\sum_{n=0}^{\infty}( a-1)x^n$

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

Now the other one.

$\displaystyle f(x) = \sum_{n=0}^{\infty}(n^3+1)x^n$

$\displaystyle f(x) = \sum_{n=0}^{\infty}(n+1)(n^2-n+1)x^n$

$\displaystyle \int f(x)~dx = F(x)$

$\displaystyle F(x) = \sum_{n=0}^{\infty}(n^2-n+1)x^{n+1}$

$\displaystyle F(x) = \sum_{n=0}^{\infty}(n^2-n)x^{n+1}+\sum_{n=0}^{\infty}x^{n+1}$

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

$\displaystyle F(x) = x^3\underbrace{\sum_{n=0}^{\infty}n(n-1)x^{n-2}}_{\text{here}}+\frac{x}{1-x}$

Now integrate that sum two times, then apply the geometric progression formula, then differentiate two times.

$\displaystyle F(x) = \frac{2x^3}{(1-x)^3}+\frac{x}{1-x}$

$\displaystyle \boxed{2}~f(x)=\frac{7x^2-2x+1}{(x-1)^4}$

Finally, $\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \frac{7x^2-2x+1}{(x-1)^4}+\frac{a-1}{1-x}$

- Jun 28th 2008, 11:41 AM #3
The other way to solve this is insipid and not that beautiful.

$\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}n^3 x^n+\sum_{n=0}^{\infty}a x^n$

$\displaystyle \sum_{n=0}^{\infty}a x^n = \frac{a}{1-x}$

$\displaystyle f = \sum_{n=0}^{\infty}n^3 x^n = x \underbrace{\sum_{n=0}^{\infty}n^3 x^{n-1}}_{g}$

$\displaystyle G = \sum_{n=0}^{\infty}n^2 x^{n} = x \underbrace{\sum_{n=0}^{\infty}n^2 x^{n-1}}_{h}$

$\displaystyle H = \sum_{n=0}^{\infty}n x^{n} = x\underbrace{\sum_{n=0}^{\infty}n x^{n-1}}_{t}$

$\displaystyle T =\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$

Now rewind it to f by differentiating and multiplying by x.

See this page for some related info:

Polylogarithm - Wikipedia, the free encyclopedia

- Jun 28th 2008, 11:49 AM #4
I am going to be lazy and skip a bunch of Latexing. But here is my approach.

I often worked these by using derivatives. A common method.

$\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}+\sum_{n=0}^{\infty}a x^{n}$

Of course, we know the second one is $\displaystyle \frac{a}{1-x}$.

Let's put it aside for the time being.

$\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}$

Let's work with the third derivative:

$\displaystyle =\frac{dx}{dx^{3}}\left[\frac{1}{1-x}\right]=\sum_{n=3}^{\infty}n(n-1)(n-2)x^{n-3}$

$\displaystyle \frac{6x^{3}}{(x-1)^{4}}=x^{3}\frac{d}{dx^{3}}\left[\frac{1}{1-x}\right]=\sum_{n=0}^{\infty}n(n-1)(n-2)x^{n}=(n^{3}-3n^{2}+2n)x^{n}$

$\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}-3\sum_{n=0}^{\infty}n^{2}x^{n}+2\sum_{n=0}^{\infty }nx^{n}$

Now, I have a few of these written down from deriving them before. Saving me the time of doing it every time.

So we get from the third derivative above:

$\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}=\frac{6x^{3}}{(x-1)^{4}}-\frac{3x(x+1)}{(x-1)^{3}}-\frac{2x}{(x-1)^{2}}=\frac{x^{3}+4x^{2}+x}{(x-1)^{4}}$

Combine with the other easier part and we joyfully get:

$\displaystyle =\frac{x^{3}+4x^{2}+x}{(x-1)^{4}}+\frac{a}{1-x}$

Which I am pretty sure is equivalent to wingless' solution, just in another form.

- Jun 28th 2008, 11:49 AM #5
Good job!

Here was my solution (pretty similar to yours)

$\displaystyle x^3\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n(n-1)(n-2)x^n=\sum_{n=0}^{\infty}n^3x^n-3\sum_{n=0}^{\infty}n^2x^n+2\sum_{n=0}^{\infty}nx^ n$

Ok, so firstly we see that

$\displaystyle \boxed{1}\quad{x\cdot\frac{d}{dx}=\sum_{n=0}^{\inf ty}nx^n}$

Next we see that

$\displaystyle x^2\cdot\frac{d^2}{dx^2}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n^2x^n-\sum_{n=0}^{\infty}nx^n$

So now using $\displaystyle \boxed{1}$ we can see that

$\displaystyle \boxed{2}\quad{x^2\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]+x\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n^2x^n}$

So now combinging $\displaystyle \boxed{1}$ and $\displaystyle \boxed{2}$

We get

$\displaystyle x^3\cdot\frac{d^3}{dx^3}\bigg[\frac{1}{1-x}\bigg]=\sum_{n=0}^{\infty}n^3x^n-3\bigg[x^2\cdot\frac{d^2}{dx^2}\bigg[\frac{1}{1-x}\bigg]+x\cdot\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]\bigg]+2x\frac{d}{dx}\bigg[\frac{1}{1-x}\bigg]$

Simplifying gives

$\displaystyle \boxed{3}\quad\frac{x(x^2+4x+1)}{(x-1)^4}=\sum_{n=0}^{\infty}n^3x^n$

And finally seeing that

$\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n=\sum_{n=0}^{\infty}n ^3x^n+a\sum_{n=0}^{\infty}x^n$

And utilizing $\displaystyle \boxed{3}$

as well as the commonly known

$\displaystyle a\sum_{n=0}^{\infty}x^n=\frac{a}{1-x}$

We get

$\displaystyle \boxed{\sum_{n=0}^{\infty}(n^3+a)x^n=\frac{x(x^2+4 x+1)}{(x-1)^4}+\frac{a}{1-x}}$

- Jun 28th 2008, 11:50 AM #6

- Jun 28th 2008, 11:51 AM #7

- Jun 28th 2008, 11:54 AM #8

- Jun 28th 2008, 11:58 AM #9
Instead of differentiating 3 times, integrate once and then divide by x, integrate, then divide by x again. Do this until you get $\displaystyle \sum_{n=0}^{\infty}x^{n}$. It will be easier to deal with. I showed this approach in my second post.

Originally Posted by**Mathstud28**

- Jun 28th 2008, 12:09 PM #10

- Jun 28th 2008, 12:19 PM #11
What? Even reading this sentence is harder than doing the calculation ;p

It's just,

$\displaystyle \sum_{n=0}^{\infty}n(n-1)x^{n-2}~\underset{\text{int}}{\to}~\sum_{n=0}^{\infty}n (n-1)x^{n}\frac{1}{n(n-1)}~\to~\sum_{n=0}^{\infty}x^{n}~\to~\frac{1}{1-x}~\underset{\text{diff}}{\to}~\frac{2}{(1-x)^3}$

- Jun 28th 2008, 12:36 PM #12

- Jun 28th 2008, 12:42 PM #13

- Jun 28th 2008, 12:43 PM #14