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I am trying to check the convergence of this series $$\sum_{n=2}^{n=\infty}\frac{ln(n)}{n\sqrt{1+n}}$$ I tried divergence test , it gives me zero ( it fails).

I tried limit comparison test with $$b_n=\sum\frac{1}{n^2}$$ but it fails (the limit gives me infinity , and bn is convergent)

I tried limit comparison with $$b_n=\sum\frac{1}{n}$$ But it fails ( the limit is zero , and bn is divergent).

I tried comparison test with (ln(n)/n) but failed also . How can I check the convergence of this series?!

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    $\begingroup$ Compare with $1/n^{4/3}$. $\endgroup$ – Daniel Fischer Dec 1 '17 at 22:01
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Let $u_n $ be the general term.

we have $$u_n\sim \frac {\ln (n)}{n^\frac 32} $$

and

$$\lim_{n\to +\infty }n^\frac 75 u_n=0,$$

because $\frac 75 <\frac 32.$

thus for great enough $n $,

$$n^\frac 75 u_n <1$$ or $$0 <u_n <\frac {1}{n^\frac 75} $$

this proves the convergence of $\sum u_n$.

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What about using Cauchy's Condensation Test?:

$$\sum_{n=1}^\infty 2^na_{2^n}=\sum_{n=1}^\infty\frac{2^n\log2^n}{2^n\sqrt{1+2^n}}=\sum_{n=1}^\infty\frac{n\log2}{\sqrt{1+2^n}}$$

But the last series converges since, for example

$$\frac1{\sqrt2}\xleftarrow[n\leftarrow\infty]{}\frac{\sqrt[n]{n\log2}}{\sqrt2\,\sqrt[n]2}=\sqrt[n]{\frac{n\log2}{\sqrt{2\cdot2^n}}}\le\color{red}{\sqrt[n]{\frac{n\log2}{\sqrt{1+2^n}}}}\le\sqrt[n]\frac{n\log2}{\sqrt{2^n}}=\frac{\sqrt[n]{n\log2}}{\sqrt2}\xrightarrow[n\to\infty]{}\frac1{\sqrt2}$$

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Creative telescoping gives another way. We have $$ \frac{\log n}{n\sqrt{n+1}}\leq 2\left[\frac{3+\log n}{\sqrt{n}}-\frac{3+\log(n+1)}{\sqrt{n+1}}\right]$$ for any $n\geq 2$, hence $$ \sum_{n=2}^{N}\frac{\log n}{n\sqrt{n+1}}\leq 2\left[\frac{3+\log 2}{\sqrt{2}}-\frac{3+\log(N+1)}{\sqrt{N+1}}\right]\leq (3+\log 2)\sqrt{2} $$ proves that the LHS is a convergent series.

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In this case the asymptotic exponent for $a_n$ is in between 1 "divergent case" and 2 "convergent case" so the series is convergent to show this you have to select an intermediate exponent for the comparison test.

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  • $\begingroup$ as n tends to infinity , do we neglect ln(n) ? I think ln(n) approaches n as n tends to infinity , so we have $$\frac{ln(n)}{n\sqrt{1+n}}\sim\frac{1}{n^{1/2}}$$ $\endgroup$ – MCS Dec 2 '17 at 23:29
  • $\begingroup$ negleting $ln(n)$, it should be $$\frac{ln(n)}{n\sqrt{1+n}}\sim\frac{1}{n^{3/2}}$$ however it can't be really negleted at all and for the comparison test you have to consider an exponent slightly greater $\endgroup$ – gimusi Dec 2 '17 at 23:35

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