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I have to study the convergence of the next series: $$\sum_{n=1}^\infty\frac{a^n}{n(1+3^n)} \text{ with }a\in R$$

Using Cauchy's theorem: $$\lim_{n\to\infty}\frac{\sqrt[\large n]{a^n}}{\sqrt[\large n]{n(1+3^n)}}=\lim_{n\to\infty}\frac{\sqrt[\large n]{a^n}}{\sqrt[\large n]{(3^n·n)+n}}\approx\lim_{n\to\infty}\frac{\sqrt[\large n]{a^n}}{\sqrt[\large n]{3^n·n}}=$$ $$\lim_{n\to\infty}\frac{\sqrt[\large n]{a^n}}{\sqrt[\large n]{3^n·n}}=\lim_{n\to\infty}\frac{\sqrt[\large n]{a^n}}{\sqrt[\large n]{3^n}·{\sqrt[\large n]{n}}}=\lim_{n\to\infty}\frac{{a}}{{3}·{\sqrt[\large n]{n}}}=\frac{a}{3}$$ Hence the limit converges when ${a}<3$ and diverges when $a>3$ and its unknown for $a=3$. Then we have to study the convergence of the next: $$\sum_{n=1}^\infty\frac{3^n}{n(1+3^n)}$$ This is the farthest that I've reached. So the questions are:

1- Is the resolution correct?

2- What can I do in order to determine the convergence in the case of $a=3$?

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3 Answers 3

up vote 4 down vote accepted

About 2. we have $$\sum_{n=1}^\infty\frac{3^n}{n(1+3^n)}=\sum_{n=1}^\infty\frac{1}{n\left(\left(\frac{1}{3}\right)^n+1\right)}$$ and $$\lim_{n\to\infty}n^{1}\frac{1}{n\left(\left(\frac{1}{3}\right)^n+1\right)}=1$$ So sice the limit is $1\neq 0$ then this series diverges.

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Perhaps I'm just being slow, but where does the $n^1$ come from in your last limit calculation? (Also, I believe you want $n\to\infty$ :) ) –  Clayton Jan 12 '13 at 18:19
    
I did the well known test for series. $\lim_{n\to\infty} n^pu_n=A$, $p=1$ and $A\neq 0$ then series is divergent. –  Babak S. Jan 12 '13 at 18:49
    
+1 Nice work. I think the test you used is just the well-known limit test: if $\,\frac{a_{n}}{b_{n}}\to r\,\,,\,\,0<r<\infty\,$ , with $\,\{a_n\}\,\{b_n\}\,$ positive sequences, then $\,\sum a_n\,$ converges iff $\,\sum b_n\,$ converges. You use this with $\,b_n=n^{-p}\,$... –  DonAntonio Jan 13 '13 at 14:01
    
@DonAntonio: Yes Don. I like your way as well, because you always never left your approaches without any details. This is very important to teach Maths. Likewise, Arturo Magidin. I love yours and Alexander Gruber's point of view in solving problems. Thanks –  Babak S. Jan 13 '13 at 14:08
    
I agree with DonAntonio: nice work! +1 –  amWhy Feb 18 '13 at 0:14

Your reasoning is correct. If the level of detail is sufficient is not mine to judge.

For the case $a=3$ you may observe that $3^n/(3^n+1)$ is bigger than $1/2$ for all $n\geq 1$ and the general term of your series is thus greater than $1/(2n)$. Conclude with the comparison theorem.

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what about the good-old ratio test?

$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{a^{n+1}}{(n+1)(3^{n+1}+1)}\frac{n(3^n+1)}{a^n}\right|=|a|\frac{n}{n+1}\frac{1}{3}\frac{1+\frac{1}{3^n}}{1+\frac{1}{3^{n+1}}}\xrightarrow [n\to\infty]{}\frac{|a|}{3}\stackrel{?}<1$$

So the series converges absolutely for $\,|a|<3\Longleftrightarrow -3<a <3\,$ , and I'll let it to you to find out whether the series converge for $\,a=\pm 3\,$ ...

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