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Let $\displaystyle \sum_{n=1}^{\infty}a_n $ be a series of real numbers that converges then prove that:

  1. the series $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n^x} $ converges uniformly on $[0, 1]$.
  2. the series $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{n+1}$ converges.

Questions: Does the series $\displaystyle \sum_{n=1}^{\infty} a_n $ need to converge absolutely in order for the conclusions to hold? May I have some hints for the first question? For the second one I tried this:

First of all I proved that the sequence goes to $0$ and then I tried the ratio test: $$\lim \left | \frac{\frac{a_{n+1}}{n+2}}{\frac{a_n}{n+1}} \right |=\left | \frac{a_{n+1}(n+1)}{a_n \left ( n+2 \right )} \right |<1 $$

because since the series $\displaystyle \sum_{n=1}^{\infty} a_n $ converges that means that $\displaystyle \lim \left | \frac{a_{n+1}}{a_n} \right |<1 $ and the other term goes to $1$. Is this OK?

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  • $\begingroup$ From the summation by parts formula, it can be determined that for a convergent sequence $\sum_{n=1}^\infty a_n$ and a monotonic sequence $b_n$, that the sequence $\sum_{n=1}^\infty a_n b_n$ converges. This is the result of part two. $\endgroup$ – Joel Mar 2 '15 at 17:02
  • $\begingroup$ You mean Abel's theorem right? That slipped my mind.. so this is easy. So, there remain only two questions. Thanks for this $\endgroup$ – Tolaso Mar 2 '15 at 17:07
  • $\begingroup$ Sure thing :) Yes I believe it is also called Abel's theorem. $\endgroup$ – Joel Mar 2 '15 at 17:15
  • $\begingroup$ By the way, for your attempt: A convergent sequence (even an absolutely convergent sequence) can have $\lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n}\right| = 1$. Take for example, $\sum 1/n^2$. All that you know, is that the ratio cannot be larger than 1. $\endgroup$ – Joel Mar 2 '15 at 17:18
  • $\begingroup$ Oh I got my mistake.. Wow! I cannot think of another test now that works here. That was the only test that pumped in my head immediately. So the other conclusion about the $\sum a_n $ does not hold either. I'm referring to $\lim |\frac{a_{n+1}}{a_n} |<1$. $\endgroup$ – Tolaso Mar 2 '15 at 17:20
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For Abel's uniform convergence test, the series converge uniformly in $\left[0,1\right]$. About your question: no, the series doesn't need to converge absolutely, we have proved the uniform convergence without this hypothesis. And no, your proof isn't right because it could be $$\left|\frac{a_{n+1}}{a_{n}}\right|\rightarrow1$$ note that $$\left|\frac{a_{n+1}}{a_{n}}\right|\rightarrow L<1$$ works if the series converge absolutely. You can prove 2. using Dirichlet's test.

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