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


closed as off-topic by José Carlos Santos, Namaste, Delta-u, Leucippus, Trevor Gunn Sep 23 '18 at 21:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Namaste, Delta-u, Leucippus, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.