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True or False? If a series $\sum_{n=1}^{\infty} a_k$ converges by either the ratio or root test, and if p > 0 is any constant, then $$\sum_{n=1}^{\infty} k^pa_k$$ converges by the same test.

My instinct is that this is true, but I am unsure as to how to go about proving it.

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  • $\begingroup$ What if $a_k=\frac1{2^k}$ and $k=2$? $\endgroup$ – Gregory Grant Feb 16 '16 at 2:20
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    $\begingroup$ What is $k$? What is $n$? Shouldn't $k$ be $n$? $\endgroup$ – Clement C. Feb 16 '16 at 2:21
  • $\begingroup$ @ClementC. obviously a typo $\endgroup$ – Gregory Grant Feb 16 '16 at 2:21
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    $\begingroup$ @GregoryGrant Yes, but I also cannot (honestly) parse the part of the question saying " then k=1 ak." (and your comment -- $p=2$, right? :) $\endgroup$ – Clement C. Feb 16 '16 at 2:22
  • $\begingroup$ @GregoryGrant Please fix your "obvious" typo - it's not obvious what you meant to say. $\endgroup$ – Zubin Mukerjee Feb 16 '16 at 2:54
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Hints: (1) What is $\lim_{k\to\infty}\frac{(k+1)^p}{k^p}$? (2) What is $\lim_{k\to\infty} (k^p)^{1/k}$?

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  • $\begingroup$ I can't read that part, it is not typeset in a way I can understand. $\endgroup$ – André Nicolas Feb 16 '16 at 2:34
  • $\begingroup$ Apologies, I have cleared up the confusion. I had accidentally duplicated two parts of the question. $\endgroup$ – BridgeSkier Feb 16 '16 at 2:37
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    $\begingroup$ I am assuming we are using the simple version of the Ratio Test.If you are using the $\limsup$ version, some small changes will have to be made. We are told that $\lim\left|\frac{a_{k+1}}{a_k}\right|\lt1$. We want to show that $\lim\left|\frac{(k+1)^pa_{k+1}}{k^pa_k}\right|\lt 1$. Since $\lim \frac{(k+1)^p}{k^p}=1$, what we want to show follows. Root Test is similar. $\endgroup$ – André Nicolas Feb 16 '16 at 2:54

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