Find values of a parameter p for which the series $\sum_{n=1}^\infty \sqrt{n}\ln^{p} \left(1+ \frac{1}{\sqrt{n}}\right)$ I was requested to find the values of the parameter p for which the following series converges:
$$\sum_{n=1}^\infty \sqrt{n} \ln^{p} \left(1+ \frac{1}{\sqrt{n}}\right)$$
I tried using Cauchy's test and the Term test, but reached a dead-end.
I also tried to use the ratio test with but it didn't seem to be helpful in this situation.
At this point we aren't allowed to use the integral test.
I would appreciate any suggestions on how to approach this problem.
 A: Hint: use the fact that
$$
x-\frac12x^2\le\log(1+x)\le x.
$$
If required it can be proved integrating the inequality
$$
1-x\le\frac1{1+x}\le1,
$$
obviously valid for $x\ge0$.
A: $$\begin{align}
\sqrt{n} \ln^{p} (1+ \frac{1}{\sqrt{n}}) 
&\overset1= \sqrt n \left( \frac{1}{\sqrt n} + O\left(\frac 1n \right)\right)^p \\
&\overset2= \sqrt n\frac{1}{(\sqrt n)^p}\left(1 + O\left(\frac{1}{\sqrt n}\right)\right) \\
&\overset3= \frac{1}{n^{p/2-1}} + O\left(\frac{1}{n^{(p-1)/2}}\right)
\end{align}$$
Explanation:

*

*Using $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3}-...$ for $|x|<1$.

*Factoring out $\frac{1}{\sqrt n}$ and using $(1+x)^p = 1 + O(x)$.

*Expanding the bracket.

A: $$\ln(1+x)\sim x;\;\text{ as }x\to 0$$
$$\ln \left(1+\frac{1}{\sqrt{n}}\right)\sim \frac{1}{\sqrt{n}};\;\text{ as }n\to\infty$$
$$\left(\ln \left(1+\frac{1}{\sqrt{n}}\right)\right)^p\sim \left(\frac{1}{\sqrt{n}}\right)^p=\frac{1}{n^{p/2}}$$
The series $$\sum_{n=1}^{\infty} \sqrt n \left(\ln \left(1+\frac{1}{\sqrt{n}}\right)\right)^p$$
converges if in the following fraction $$\sqrt n\frac{1}{n^{p/2}}=\frac{1}{n^{p/2-1/2}}$$
we have p_series test $$\frac{p}{2}-\frac{1}{2}>1\to p>3$$
A: We have that
$$\frac{\sqrt{n} \ln^{p} \left(1+ \frac{1}{\sqrt{n}}\right)}{\left(\frac1{\sqrt{n}}\right)^{p-1}}=\left(\frac{\ln \left(1+ \frac{1}{\sqrt{n}}\right)}{\frac1{\sqrt{n}} }\right)^p \to 1$$
therefore by limit comparison test the given series converges if and only if the series $\sum \left(\frac1{\sqrt{n}}\right)^{p-1}$ converges that is for $$(p-1)/2>1 \iff p>3$$
