# Values that make series converge

• Jun 10th 2009, 09:43 PM
utopiaNow
Values that make series converge
Find the values for $x \in \mathbb{R}$ for which the following series converge:

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

Attempted solution:
I tried to use the root test to evaluate $\displaystyle\lim_{n\to\infty} |a_n|^{1/n}$ and found that it is equal to $\displaystyle\lim_{n\to\infty} |x(1 - x^n)|^{1/n}$. I got stuck here so I tried to use the ratio test and got to $\displaystyle\lim_{n\to\infty} |a_n|^{1/n} = \displaystyle\lim_{n\to\infty} \frac{x(1 - x^{n+1})}{1 - x^n}$. And I'm stuck here for this method as well.

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

Attempted solution:
I thought the only useful test would be the ratio test. So I tried it and got
$\displaystyle\lim_{n\to\infty} \left|\frac{b_{n + 1}}{b_n}\right| = \displaystyle\lim_{n\to\infty} \frac{2n+ 1}{(n + 1)^3}(x - e)$ however then I found $\displaystyle\lim_{n\to\infty} \frac{2n+ 1}{(n + 1)^3} = 0$ which implies all values $x$ make this series converge, but that doesn't seem right.

Any suggestions would be appreciated. Thanks in advance.
• Jun 10th 2009, 10:41 PM
chisigma
The series can be 'splitted' as...

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

... and it becomes the sum of two series both converging for $|x|<1$ ...

Kind regards

$\chi$ $\sigma$
• Jun 11th 2009, 12:16 AM
flyingsquirrel
Quote:

Originally Posted by utopiaNow
2. $\displaystyle\sum_{n=1}^{\infty}\left[\frac{(2n)!}{n(n!)^2}\right](x - e)^n$

...which implies all values $x$ make this series converge, but that doesn't seem right.

You're correct, it converges for all values of $x$.
Quote:

Originally Posted by chisigma
The series can be 'splitted' as...

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

... and it becomes the sum of two series both converging for $|x|<1$ ...

But it doesn't mean that $\sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}$ converges iff both $\sum_{n=1}^{\infty} \frac{x^{n}}{n}$ and $\sum_{n=1}^{\infty} \frac{x^{2n}}{n}$ converge. For example for $x=1$ the first series does converge but the two others don't. It remains to be checked that $\sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}$ diverges when $x=-1$ or $|x|>1$.

I think that using the ratio test is simpler.

$\frac{\frac{x^{n+1} (1-x^{n+1})}{n+1}}{\frac{x^{n} (1-x^{n})}{n}}
=\frac{x(1-x^{n+1})}{1-x^n}\cdot \frac{n}{n+1}$

If $|x|<1$, $x^n\xrightarrow[n\to\infty]{}0$ hence
$\lim_{n\to\infty}\frac{x(1-x^{n+1})}{1-x^n}\frac{n}{n+1}=x\cdot \lim_{n\to\infty}\frac{1-x^{n+1}}{1-x^n}\cdot \lim_{n\to\infty}\frac{n}{n+1}=x\cdot \frac{1-0}{1-0} \cdot 1 =x$

and as $|x|<1$, the series converges. Similarly one can show that if $|x|>1$ the series diverges. To check if the series converges when $|x|=1$ simply substitute $x=1$ or $x=-1$ :

• If $x=1$,
$\sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} 0.$
Is that a convergent series ? (Rofl)
• If $x=-1$,
$\sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} \frac{(-1)^{n}\cdot (1-(-1)^{n})}{n}=\sum_{n=1}^{\infty} \frac{-1}{2n-1}$
because $1-(-1)^{n}=0$ when $n$ is even. Does this series converge ?
• Jun 11th 2009, 12:39 AM
utopiaNow
Quote:

Originally Posted by flyingsquirrel
• If $x=-1$,
$\sum_{n=1}^{\infty} \frac{x^{n}\cdot (1-x^{n})}{n}=\sum_{n=1}^{\infty} \frac{(-1)^{n}\cdot (1-(-1)^{n})}{n}=\sum_{n=1}^{\infty} \frac{-1}{2n-1}$
because $1-(-1)^{n}=0$ when $n$ is even. Does this series converge ?

I believe it is divergent, diverging to $-\infty$. Would the formal reason be that it diverges because it is a general harmonic series?
• Jun 11th 2009, 12:42 AM
flyingsquirrel
Quote:

Originally Posted by utopiaNow
Would the formal reason be that it diverges because it is a general harmonic series?

Yes. (Nod)