# Series problems

• Dec 14th 2011, 10:26 PM
maoro
Series problems
#1
Find the radius of convergence and the interval of convergence of the series:
$\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{nln(1+n)}$
I have attempted this problem as following
By root test
$\displaystyle lim_{n\to\infty}\sqrt[n]{|\frac{x^n}{nln(1+n)}|}$
$\displaystyle =lim_{n\to\infty}|x|\sqrt[n]{|\frac{1}{nln(1+n)}|}$
$\displaystyle =lim_{n\to\infty}|x|\frac{1}{|\sqrt[n]{ln(1+n)}|}$
$\displaystyle =\frac{|x|}{lim_{n\to\infty}\sqrt[n]{ln(1+n)}}$
$\displaystyle =\frac{|x|}{\sqrt[n]{lim_{n\to\infty}ln(1+n)}}$
$\displaystyle =\frac{|x|}{\sqrt[n]{lim_{n\to\infty}ln(1+n)}}$
$\displaystyle =\frac{|x|}{\sqrt[n]{ln(lim_{n\to\infty}(1+n))}}$
$\displaystyle =|x|$

When $\displaystyle x=1$,$\displaystyle \sum_{n=0}^{\infty}\frac{1}{nln(1+n)}$ converges.
When $\displaystyle x=-1$,$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{nln(1+n)}$converges.
Therefore, $\displaystyle R=1$ and $\displaystyle I=[-1,1]$

Is it correct?

#2
Assume that the positive series $\displaystyle \sum_{n=0}^{\infty} a_n$ is convergent.
Prove that the series $\displaystyle \sum_{n=0}^{\infty} ({a_n}^4+\frac{1}{n^2})$ is also convergent.
• Dec 14th 2011, 10:30 PM
Chris11
Re: Series problems
For a power series, you do not have to prove convergance using root tests, because you know that the series converges within its radius of convergance, and this radius is defined whenever the limit lim a_n/a_n+1 exists. So compute the limit, then test the end points.
• Dec 14th 2011, 10:57 PM
maoro
Re: Series problems
Quote:

Originally Posted by Chris11
For a power series, you do not have to prove convergance using root tests, because you know that the series converges within its radius of convergance, and this radius is defined whenever the limit lim a_n/a_n+1 exists. So compute the limit, then test the end points.

But is that true for using root test to prove it and is my computation true?
• Dec 15th 2011, 02:19 AM
Prove It
Re: Series problems
Quote:

Originally Posted by Chris11
For a power series, you do not have to prove convergance using root tests, because you know that the series converges within its radius of convergance, and this radius is defined whenever the limit lim a_n/a_n+1 exists. So compute the limit, then test the end points.

Surely you mean the radius of convergence is defined wherever \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \end{align*}...
• Dec 15th 2011, 03:17 AM
FernandoRevilla
Re: Series problems
Quote:

Originally Posted by maoro
#2Assume that the positive series $\displaystyle \sum_{n=0}^{\infty} a_n$ is convergent.
Prove that the series $\displaystyle \sum_{n=0}^{\infty} ({a_n}^4+\frac{1}{n^2})$ is also convergent.

As $\displaystyle a_n\to 0$ there exists $\displaystyle n_0\in\mathbb{N}$ such that $\displaystyle a_n^4\leq a_n$ for all $\displaystyle n\geq n_0$ . Hence, $\displaystyle 0\leq a_n^4+\frac{1}{n^2}\leq a_n+\frac{1}{n^2}$ and $\displaystyle \sum_{n=1}^{\infty}\left(a_n+\frac{1}{n^2}\right)$ is convergent. So, ...

P.S. Surely there is a typo in your post: $\displaystyle n=1$ instead of $\displaystyle n=0$ .
• Dec 15th 2011, 04:15 AM
maoro
Re: Series problems
Here is another problem is need help:
Use the ratio test to prove that the series $\displaystyle \sum_{n=1}^{\infty} \sin{\frac{\pi}{n}$ diverges
• Dec 15th 2011, 04:52 AM
FernandoRevilla
Re: Series problems
$\displaystyle \lim_{n\to \infty}\frac{\sin (\pi /n)}{\pi /n}=1\neq 0$ , $\displaystyle \sin (\pi/n)\geq 0$ for all $\displaystyle n$ , and $\displaystyle \sum_{n=1}^{\infty}\pi/n$ is divergent, so the given series is divergent.
• Dec 15th 2011, 04:55 AM
Prove It
Re: Series problems
Quote:

Originally Posted by maoro
Here is another problem is need help:
Use the ratio test to prove that the series $\displaystyle \sum_{n=1}^{\infty} \sin{\frac{\pi}{n}$ diverges

The ratio test is inconclusive because \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\sin{ \left( \frac{\pi}{n + 1} \right) } }{ \sin{ \left( \frac{\pi}{n} \right) } } \right| = 1 \end{align*}...
• Dec 15th 2011, 04:58 AM
maoro
Re: Series problems
Quote:

Originally Posted by Prove It
The ratio test is inconclusive because \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\sin{ \left( \frac{\pi}{n + 1} \right) } }{ \sin{ \left( \frac{\pi}{n} \right) } } \right| = 1 \end{align*}...

ya, i also got this result and can't prove the desired divergence.
• Dec 15th 2011, 08:56 AM
FernandoRevilla
Re: Series problems
Quote:

Originally Posted by maoro
ya, i also got this result and can't prove the desired divergence.

For that reason I applied the Limit Comparison Test: if $\displaystyle \sum a_n$ and $\displaystyle \sum b_n$ are series with positive terms and the limit $\displaystyle \lim_{n\to \infty}\frac{a_n}{b_n}$ exists and is nonzero, then either both $\displaystyle \sum a_n$ and $\displaystyle \sum b_n$ converge, or else both diverge.
• Dec 15th 2011, 12:39 PM
Chris11
Re: Series problems
@proveit I was just telling him what the radius was.