# Does the series $\sum_{ n = 0}^{\infty } \frac{1}{\sqrt{n}}\ln\bigl(1+\frac{1}{\sqrt{n}}\bigr)$ converge? [closed]

$$\sum_{ n = 0}^{\infty } U_n = \frac{1}{\sqrt{n}}\ln\left(1+\frac{1}{\sqrt{n}}\right)$$ 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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• Hint: Note that this converges if and only if the sequence with $\sqrt{n}$ replaced by $n$ converges. – Mees de Vries Sep 21 '18 at 10:24
• Are you talking about a series or a sequence? The sequence $\{U_n\}$ is convergent but the series $\sum \{U_n\}$ is divergent. – Kavi Rama Murthy Sep 21 '18 at 10:25
• i am talking about a serie – KEVIN DLL Sep 21 '18 at 10:27

## 3 Answers

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.

• By limit comparison test, it should be added. – gimusi Sep 21 '18 at 10:52
• @gimusi: Done, mylord! – Bernard Sep 21 '18 at 11:10
• That’s perfect now! Cheers :) – gimusi Sep 21 '18 at 11:12

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.

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.

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