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I am checking for convergence of

$$\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$$ First, we notice $${n+1 \over n-1}= {1+{2 \over n-1}}$$ Then, we use $\ln(1+x) \le x$ and get $$\ln\left({1 + {2 \over n-1}}\right) \le {2 \over n-1}$$ $${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1}$$ We now use $n-1 > {n \over 2}$ $${1 \over \sqrt{n}}\ln{n+1 \over n-1} \le {1 \over \sqrt{n}}{2 \over n-1} < {1 \over \sqrt{n}}{2 \over {n \over 2}} = 4{1 \over n\sqrt{n}}$$ $\sum_{n=2}^{\infty}4{1 \over n\sqrt{n}}$ converges and so does $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$. Is this correct? Alternatively, is there a quicker way?

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    $\begingroup$ This is correct, and I see no better way. Well done! $\endgroup$ – 5xum Dec 1 '16 at 8:15
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You are right. Alternatively you can use the asymptotic comparison test by noting that $${1 \over \sqrt{n}}\ln\left({n+1 \over n-1}\right)= {1 \over \sqrt{n}}\ln\left({1+{2 \over n-1}} \right)\sim \frac{2}{n\sqrt{n}}=\frac{2}{n^{3/2}}$$ where we used that $\ln(1+x)\sim x$ when $x\to 0^+$.

It is similar to your approach but you do not deal with inequalities.

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    $\begingroup$ You don't need $0^+$ , only $0$.friendly. $\endgroup$ – hamam_Abdallah Dec 1 '16 at 8:27
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If we use the well-known equivalence,

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

we get that

$$\ln(1+\frac{2}{n-1})\sim\frac{2}{n-1}\sim\frac{2}{n}\;\;(n\to +\infty)$$

$$\implies \frac{1}{\sqrt{n}}\ln(\frac{n+1}{n-1})\sim\frac{2}{n^\frac{3}{2}}\;(n\to+\infty)$$

the general terms are positive and equivalent, the series are both convergent since $\frac{3}{2}>1$.

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