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$$\sum_{ n = 0}^{\infty } U_n = \frac{1}{\sqrt{n}}\ln\left(1+\frac{1}{\sqrt{n}}\right)$$ I was trying to resolve it by any method convergence, but I could not show if the series converges or diverges.

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  • 1
    $\begingroup$ Hint: Note that this converges if and only if the sequence with $\sqrt{n}$ replaced by $n$ converges. $\endgroup$ – Mees de Vries Sep 21 '18 at 10:24
  • $\begingroup$ Are you talking about a series or a sequence? The sequence $\{U_n\}$ is convergent but the series $\sum \{U_n\}$ is divergent. $\endgroup$ – Kavi Rama Murthy Sep 21 '18 at 10:25
  • $\begingroup$ i am talking about a serie $\endgroup$ – KEVIN DLL Sep 21 '18 at 10:27
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Since it is a series with positive terms, the simplest is to use equivalents (by limits comparison test):

We know that $\;\ln(1+x)\sim_0 x$, so we deduce an equivalent for the general term of this series: $$\frac1{\sqrt n}\,\log\Bigl(1+\frac1{\sqrt n}\Bigr)\sim_{n\to\infty}\frac1{\sqrt n}\,\frac1{\sqrt n}=\frac 1n,$$ which is the general term of the (divergent) harmonic series.

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  • $\begingroup$ By limit comparison test, it should be added. $\endgroup$ – gimusi Sep 21 '18 at 10:52
  • $\begingroup$ @gimusi: Done, mylord! $\endgroup$ – Bernard Sep 21 '18 at 11:10
  • $\begingroup$ That’s perfect now! Cheers :) $\endgroup$ – gimusi Sep 21 '18 at 11:12
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Compare with $\sum \frac 1 n$. The series is divergent. Hint: $\log (1+x) \geq \frac 1 2 x$ for $x>0$ and sufficiently small.

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Note that

$$\frac1{\sqrt n}\log \left(1+\frac1{\sqrt n}\right)\sim \frac1{\sqrt n}\frac1{\sqrt n} =\frac1n$$

then refer to limit comparison test.

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  • $\begingroup$ can you explain it ? i don't understand $\endgroup$ – KEVIN DLL Sep 21 '18 at 10:37
  • $\begingroup$ Recall that as $x\to 0$ we have $\log(1+x)\sim x$ therefore $$\frac1{\sqrt n}\log \left(1+\frac1{\sqrt n}\right)\sim \frac1{\sqrt n}\frac1{\sqrt n} =\frac1n$$ $\endgroup$ – gimusi Sep 21 '18 at 10:50
  • $\begingroup$ Do you know limit comparison test? $\endgroup$ – gimusi Sep 21 '18 at 10:50
  • $\begingroup$ You can find a reference here LCT and note that it is necessary refer to it for a rigorous proof. $\endgroup$ – gimusi Sep 21 '18 at 10:54

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