1. ## Sum

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.

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

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.

Polylogarithm - Wikipedia, the free encyclopedia

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.

5. Originally Posted by wingless
$\displaystyle \sum_{n=0}^{\infty}(n^3+a)x^n = \sum_{n=0}^{\infty}(n^3+1)x^n+\sum_{n=0}^{\infty}( a-1)x^n$

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

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

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

6. Hey Mathstud. Looks me and you were thinking a like. Good to see we checked out.

7. Originally Posted by galactus
Hey Mathstud. Looks me and you were thinking a like. Good to see we checked out.

8. Originally Posted by wingless
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.

Polylogarithm - Wikipedia, the free encyclopedia
Insipid and not beautiful eh?

9. Originally Posted by galactus
$\displaystyle \sum_{n=0}^{\infty}n^{3}x^{n}$

Let's work with the third derivative:

(...)

$\displaystyle =\frac{x^{3}+4x^{2}+x}{(x-1)^{4}}+\frac{a}{1-x}$
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
Insipid and not beautiful eh?
It's not as creative as the other solution, but it works for any $\displaystyle \sum_{n=0}^{\infty}n^k x^n$

10. Originally Posted by wingless
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.

It's not as creative as the other solution, but it works for any $\displaystyle \sum_{n=0}^{\infty}n^k x^n$
I think yours was much more work though, creatively disguised under

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

11. Originally Posted by Mathstud28
I think yours was much more work though, creatively disguised under
Now integrate that sum two times, then apply the geometric progression formula, then differentiate two times.
What? Even reading this sentence is harder than doing the calculation ;p

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

12. Originally Posted by wingless
What? Even reading this sentence is harder than doing the calculation ;p

It's just,
$\displaystyle \sum_{n=0}^{\infty}n(n-1)x^{n-2}~\underset{\text{int}}{\to}~\sum_{n=0}^{\infty}n (n-1)x^{n}\frac{1}{n(n-1)}~\to{\color{red}{\cdots}}~\sum_{n=0}^{\infty}x^ {n}~\to~\frac{1}{1-x}~\underset{\text{diff}}{\to}{\color{red}{\cdots} }~\frac{2}{(1-x)^3}$
...

But anways, even if your solution was more beautiful (which it is ), don't say another person's solution isn't as beautiful..even though I know you had no intention of it coming off that way...all math is equally beautiful!....except statistics

13. I said "not beautiful" for my own solution. Come on, do you really think I'm that rude?

14. Originally Posted by wingless
I said "not beautiful" for my own solution. Come on, do you really think I'm that rude?
No ...I'm just a wuss