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

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.