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$$\sum_{k=1}^\infty\left(\frac k{k+1}\right)^{k^2}$$

Determine whether or not the following series converge. I am not sure what test to use. I am pretty sure I am unable to use ratio test. Maybe comparison or Kummer, or Raabe. However I am not sure how to start it.

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  • $\begingroup$ You have an exponent with $k$. Your instinct should be to remove it with the root test. $\endgroup$ – Simply Beautiful Art Apr 15 at 2:38
  • $\begingroup$ Since its squared, do I take the k^2 root $\endgroup$ – radius Apr 15 at 2:40
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    $\begingroup$ Does the root test say you take the $k^2$-th root? $\endgroup$ – Simply Beautiful Art Apr 15 at 2:44
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    $\begingroup$ For $k\ge 1$, we have$$\left(\frac{k}{k+1}\right)^{k^2}\le e^{-k/2}$$ $\endgroup$ – Mark Viola Apr 15 at 3:14
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By Root test, $$\limsup_{n \to \infty} \sqrt[n]{\left(\frac{n}{n+1}\right)^{n^2}}=\limsup_{n \to \infty} {\left(\frac{n}{n+1}\right)^{n}}=\limsup_{n \to \infty} {\left({1+\frac{1}{n}}\right)^{-n}}=\frac{1}{e}<1$$

So your series converges!

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  • $\begingroup$ How did you know to take the supremum $\endgroup$ – radius Apr 15 at 2:51
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    $\begingroup$ See the wiki link! $\endgroup$ – Chinnapparaj R Apr 15 at 2:54
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Hint: $$\left( \frac{k}{k+1} \right)^k \sim e^{-1} $$

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$$a_k=\left(\frac k{k+1}\right)^{k^2}\implies \log(a_k)=k^2 \log\left(\frac k{k+1}\right)$$

$$\log(a_{k+1})-\log(a_k)=(k+1)^2 \log \left(\frac{k+1}{k+2}\right)-k^2 \log \left(\frac{k}{k+1}\right)$$ Using Taylor expansions for large $k$ $$\log(a_{k+1})-\log(a_k)=-1+\frac{1}{3 k^2}+O\left(\frac{1}{k^3}\right)$$ $$\frac {a_{k+1}}{a_k}=e^{\log(a_{k+1})-\log(a_k)}=\frac 1 e \left(1+\frac{1}{3 k^2}+O\left(\frac{1}{k^3}\right)\right)\to \frac 1 e $$

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