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Does $\Sigma_{n=1}^{\infty} \dfrac {\sqrt {2n-1}~ \log (4n+1)} {n(n+1)}$ converge?

Attempt: I have been trying to use the comparison test for a while, but I can't find a suitable comparator.

For Example, since, $\Sigma_{n=1}^{\infty} \dfrac {1}{n(n+1)}$ converges, I was thinking of an expression involving the above to use as a comparator. But can not find one.

Please guide me on how to move ahead.

Thank you very much for your help.

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First using the L'Hôpital's rule we see that

$$\ln(n)=_\infty o(n^\alpha),\quad \forall \alpha>0$$ Now we have

$$\frac {\sqrt {2n-1}~ \log (4n+1)} {n(n+1)}\sim_\infty \sqrt 2\frac{\ln(4n)}{n^{3/2}}=o\left(\frac1{n^{3/2-\alpha}}\right)$$ so it suffices to choose $\alpha>0$ such that $\frac32-\alpha>1$ to conclude that the given series is convergent.

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  • $\begingroup$ $\sqrt 2\frac{\ln(4n)}{n^{3/2}}=o\left(\frac1{n^{3/2-\alpha}}\right)$. Could you please explain the steps after that? $\endgroup$ – MathMan Jan 22 '15 at 13:34
  • $\begingroup$ If you choose $\alpha>0$ st $\frac{3}{2}-\alpha>1$ then $\sum \frac{1}{n^{3/2-\alpha}}$ converges so your series will also converge. $\endgroup$ – Bhauryal Jan 22 '15 at 18:54
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For large $n$ $$\frac{\sqrt{2n-1} \hspace{1mm}\ln(4n+1) }{ n(n+1) }\approx \frac{\sqrt{2n} \hspace{1mm}\ln(4n) }{ n(n+1) }=\sqrt{2}\frac{\ln4n}{\sqrt{n}(n+1)}=\sqrt{2} \hspace{1mm} \ln4\frac{1}{\sqrt{n}(n+1)}+\sqrt{2}\frac{\ln n}{\sqrt{n}(n+1)}$$ The first term is the term from a convergent series.(Hint:- It is dominated by a convergent $p$-series). Now consider the series $\sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n}n}$. We are done if we could show that this series converges. Integral test seems to be efficient in this case:- \begin{equation} \begin{aligned} \lim_{n \to \infty}\int_1^{n}\frac{\ln x}{x\sqrt{x}} dx&=\lim_{n \to \infty}\int_0^{\ln n}ue^{-\frac{u}{2}}du \qquad (\mbox{where $u=\ln x$} )\\ &=\int_0^{\infty}ue^{-\frac{u}{2}}du \\ &=4 \end{aligned} \end{equation}Thus $\sum_{n=1}^{\infty}\frac{\ln n}{n\sqrt{n}}$ converges. Since $\frac{\ln n}{\sqrt{n} (n+1)}<\frac{\ln n}{n\sqrt{n}}$, so by Comparison test the series $\sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} (n+1)}$ converges. Ultimately, your series converges.

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