I need help with the following:

We are given the series $\sum_1^\infty \frac{a_n}{5^n}$ and $\sum_1^\infty \frac{(-1)^na_n}{5^n}$ . We also know that the first series converges while the second one diverges. Now we must answer these questions:

1) is series $\sum_1^\infty \frac{a_n}{5^n}$ a)absolutely convergent or b)conditionally convergent?

2) is the series $\sum_1^\infty \frac{a_n}{4^n}$ a)absolutely convergent, b)conditionally convergent, c) divergent

3) is the series $\sum_1^\infty \frac{a_n}{6^n}$ a)absolutely convergent, b)conditionally convergent, c) divergent

4) What is the radius of convergence of is the series $\sum_1^\infty (n+1)a_nx^n$

1b) seems obvious and I can easily see that 2a) is wrong by the comparison test. But how can I find the exact answer? I tried using the series test, but it doesn't give me information about conditional divergance.

Any help would be appreciated

  • $\begingroup$ Got something from an answer below? $\endgroup$ – Did Jul 29 '12 at 14:05

Re 2., if the sequence $\left(\frac{a_n}{4^n}\right)_{n\geqslant0}$ is bounded, then $|a_n|\leqslant c\cdot4^n$ for some finite $c$, for every $n\geqslant0$, hence $\left|\frac{a_n}{5^n}\right|\leqslant c\cdot\left(\frac45\right)^n$ and $\sum\limits_n\frac{a_n}{5^n}$ is absolutely convergent. This proves that $\left(\frac{a_n}{4^n}\right)_{n\geqslant0}$ is not bounded hence $\sum\limits_n\frac{a_n}{4^n}$ diverges (2.c).

Re 3., $\sum\limits_n\frac{a_n}{5^n}$ converges hence $\left(\frac{a_n}{5^n}\right)_{n\geqslant0}$ is bounded, hence $\left|\frac{a_n}{6^n}\right|\leqslant c\cdot\left(\frac56\right)^n$ for some finite $c$, for every $n\geqslant0$, and $\sum\limits_n\frac{a_n}{6^n}$ converges absolutely (3.a).

Re 4., the radius of convergence of the series $\sum\limits_n(n+1)a_nx^n$ and $\sum\limits_na_nx^n$ are the same (always). The divergence result of 2. holds for every $x\gt\frac15$ instead of $\frac14$ and the absolute convergence result of 3. holds for every $x\lt\frac15$ instead of $\frac16$. Hence the radius of convergence is $R=\frac15$.


Hint: if $$\limsup_{n\to\infty}\left(\frac{|a_n|}{5^n}\right)^{1/n}< 1$$ then the series $\sum_1^\infty \frac{a_n}{5^n}$ would converge absolutely.

  • $\begingroup$ So I can take $(\frac{|a_n|}{5})^n$ and it would be bigger or equal to 1. Therefore if we take $(\frac{|a_n|}{6})^n = (\frac{|a_n|}{5} \frac{5}{6})^n$ - its limit would be 0 and therefore the third series is absolutely convergent, right? If this is correct, then the second one is not absolutely convergent but how can I prove conditional convergence when I can't use the Leibniz critiria? $\endgroup$ – LLT Jun 28 '12 at 14:58
  • $\begingroup$ @LLT $|a_n|$ isn't taken to the $n$th degree. And if the limit is greater than one then.... $\endgroup$ – Andrew Jun 28 '12 at 15:17
  • $\begingroup$ Damn, my bad. But then we get that all three series have the same limit( because $(\frac{5}{6})^\frac{1}{n}$ has limit 1) or I am wrong again? This isn't really my calculus day;/ But what if the limit is 1? Then we can't say anything? $\endgroup$ – LLT Jun 28 '12 at 15:25
  • $\begingroup$ @LLT about $a_n/n^5$ no, but about the others we can. $\endgroup$ – Andrew Jun 28 '12 at 15:58
  • $\begingroup$ Thanks for the help @Andrew, but unfortunately I couldn't figure out the exact prove, so I can only hope this one isn't on the exam tomorrow :) Some friend says he proved that the ratio test returns $\frac{a_n+1}{a_n}$ is equal to 1 but I don't think that's true(because he uses the ratio test in a strange way). Still, if anyone can help with the answer, it'd be appreciated, because I am curious what's the catch that I'm missing :) $\endgroup$ – LLT Jun 28 '12 at 18:41

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