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$$\sum\limits_{k=1}^{\infty}\prod\limits_{j=1}^{k}\frac{4j^2-6j+3}{2j(2j+1)}t^{2k+1}$$ I could prove using ratio test that if $|t|<1$ this above series converges. But wolfram alpha shows when $|t|=1$ this series still converges. I am stuck here. Ratio test is inconclusive. Any help would be appreciated.

mudok
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  • Looks like a job for Raabe. – Daniel Fischer Aug 23 '17 at 17:51
  • @DanielFischer Thanks I will try this beautiful test. Intriguingly wolfram alpha says "by comparison test" this converges when $|t|=1$. I am still searching for a series which dominates this . – mudok Aug 23 '17 at 17:56
  • $$\sum_{k = 1}^{\infty} \frac{1}{k^2}$$ dominates it, for example. – Daniel Fischer Aug 23 '17 at 17:58
  • Im sorry sir.. I cant see..:( – mudok Aug 23 '17 at 18:01
  • Comes out of Raabe's test. We have $\frac{a_k}{a_{k-1}} = 1 - \frac{2}{k} + O(k^{-2})$, hence the sequence $a_k$ behaves asymptotically like $\frac{c}{k^2}$. – Daniel Fischer Aug 23 '17 at 18:08
  • I applied raabe's test and found that $k(\frac{a_{k+1}}{a_k}-1)\to 2$ as $k\to\infty$. Therefore the series converges. But could not understand your argument. How $a_k=\mathcal{O}(1/k^2)$? – mudok Aug 23 '17 at 18:15
  • Take logarithms. $\log \frac{a_k}{a_{k-1}} = - \frac{2}{k} + O(k^{-2})$, so summing that up yields $\log a_k - \log a_1 = -2\log k + O(1)$, and hence $a_k = \frac{a_1}{k^2}\cdot e^{O(1)}$. – Daniel Fischer Aug 23 '17 at 18:39
  • Fantastic argument Sir. Many thanks..:) – mudok Aug 23 '17 at 18:49

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hint

For $|t|=1$, take logarithm of the product and find an equivalent to $$\ln ( \frac {4j^2-6j+3}{2j (2j+1)} )=$$ $$\ln (1+\frac {-8j+3}{2j (2j+1)} )$$ using

$$\ln (1+X)\sim X \;\;(X\to 0) $$

hamam_Abdallah
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