Domain of convergence of power series-2

Question

Here is the series:

$$ \sum_{n=2}^\infty(-1)^n\frac{(x-3)^n}{(\sqrt[n]{n}-1)n} $$

What is the interval of convergence? I tried using root test and ratio test but finding the limit from thereon is quite difficult.

From root test: $$ \lim_{n->\infty}\frac{|x-3|}{(\sqrt[n]{\sqrt[n]{n}-1)n}} $$

How about $$n^{1/n} \sim 1+{\frac {\ln \left( n \right) }{n}}+{\frac { \left( \ln \left( n \right) \right) ^{2}}{2{n}^{2}}}+O \left( {n}^{-3} \right) $$ — GEdgar, Oct 02 '13 at 12:40
I don't think its a very solid way of doing it because it leads to 0^0?? — John, Oct 02 '13 at 12:46
For inside the interval of convergence how about applying GEdgar suggestion together with alternating series test? — Maesumi, Oct 02 '13 at 12:48

Arash · Accepted Answer · 2013-10-02T13:33:59.210

0

Let's go for the root test: $$ r=\displaystyle\lim_{n\to\infty}\frac{|(x-3)|}{\sqrt[n]{(\sqrt[n]{n}-1)n}} $$ So we have to calculate the following limit: $$ \lim_{n\to\infty}{\sqrt[n]{(\sqrt[n]{n}-1)n}} $$ Form here on, there is a mundane way to show that the limit is one. The direct path is using multiple Hopital. But we give another proof:

$$ \lim_{n\to\infty}{\sqrt[n]{(\sqrt[n]{n}-1)n}}=\lim_{n\to\infty}{\sqrt[n]{n}}{\sqrt[n]{\sqrt[n]{n}-1}} $$ It can be shown that ${\sqrt[n]{n}}$ goes to 1 as $n\to\infty$. For the other limit, see that: $$ \frac{\ln n}{n}\leq \sqrt[n]{n}-1\leq n\implies \sqrt[n]{\frac{\ln n}{n}}\leq \sqrt[n]{\sqrt[n]{n}-1}\leq \sqrt[n]{n} $$ Now you can show that both limits of LHS and RHS go to 1 as $n\to\infty$ and therefore the limit in the middle exists and is equal to one and hence the final limit exists and is 1.

In any case the domain of convergence is: $$ {|(x-3)|}<1 $$ with the point $x=4$.

Remark 1: $$ \sqrt[n]{\frac{\ln n}{n}}=L\implies \frac{1}{n}\ln(\frac{\ln n}{n}) =\ln L\\ \lim_{n\to\infty} \frac{\ln({\ln n})-\ln({n})}{n}=0\implies L=1. $$

Remark 2: If $x-3=-1$, the series becomes: $$ \sum_{n=2}^\infty(-1)^{2n}\frac{1}{(\sqrt[n]{n}-1)n}=\sum_{n=2}^\infty\frac{1}{(\sqrt[n]{n}-1)n} $$ Such series with positive non-increasing terms converges iff $\sum 2^na_{2^n}$ converges. Hence: $$ \sum_{n=2}^\infty\frac{2^n}{(\sqrt[n]{2^n}-1)2^n}=\sum_{n=2}^\infty {1}\to\infty $$ Therefore it diverges.

For the case were $x-3=1$, we get an alternating series with non-increasing absolute value of its terms and according to the alternating series test it converges.

edited Oct 02 '13 at 13:33

answered Oct 02 '13 at 12:51

Arash

11,131

The step to prove that the limit is 1 using L'hopitals is really long and infeasible – John Oct 02 '13 at 12:52
There are easier ways as mentioned in the comments. Know that ${\sqrt[n]{n}}$ goes to 1 as $n\to\infty$. – Arash Oct 02 '13 at 12:58
I know that...thanks for your alternative – John Oct 02 '13 at 13:04
I added something to the answer. – Arash Oct 02 '13 at 13:05
What did you add?--Its by the Squeeze Theorem – John Oct 02 '13 at 13:06
Explain how you ruled out cases $|x-3|=1$. – GEdgar Oct 02 '13 at 13:08
I think a proper proof that the LHS goes to 1 is also needed since it is an inderminate form – John Oct 02 '13 at 13:09
@GEdgar, Very good point indeed. Added to the answer. – Arash Oct 02 '13 at 13:34
The OP did say interval of convergence, so I will let you off the hook concerning the rest of the circle $|x-3|=1$ in the complex plane. – GEdgar Oct 02 '13 at 15:41
Yes, I had the same impression from OP. – Arash Oct 02 '13 at 16:05

Domain of convergence of power series-2

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