Convergence series with natural logarithm Let $p \in \mathbb{R}$ and $p>0$. When the series
$ \sum_{n=1}^{\infty} \frac{\ln(1+n^p)}{n^p}$
is convergence (dependent from parametr $p$) ?
Of course, $\lim_{n \to \infty} \frac{\ln(1+n^p)}{n^p} = 0$. It is true, because $p>0$.
Next, I tried different convergence tests (especially D'Alembert's criterion, comparison test,   Dirichlet's test) but without success.
I will grateful for you hints and help.
 A: First note that a positive series either converges or diverges (to $+\infty$).
Now, let $0 < p \leq 1$. Then, $$\sum_{n=1}^{\infty} \frac{\log(1+n^p)}{n^p} \geq \sum_{n=1}^{\infty} \frac{\log(1+n^p)}{n} \geq \sum_{n=k}^{\infty} \frac{1}{n},$$where $k$ is the smallest positive integer such that $\log(1+k^p) \geq 1$. Since the lower bound diverges, the original series diverges.
Now, let $p = 1+\epsilon$ for some $\epsilon>0$. We have $$\sum_{n=1}^{\infty} \frac{\log(1+n^p)}{n^p} \leq \sum_{n=1}^{k-1} \frac{\log(1+n^p)}{n^p} + \sum_{n=k}^{\infty} \frac{n^{\epsilon/2}}{n^{1+\epsilon}} = \sum_{n=1}^{k-1} \frac{\log(1+n^p)}{n^p} + \sum_{n=k}^{\infty} \frac{1}{n^{1+\epsilon/2}},$$ where this $k$ is the smallest positive integer such that $\log(1+k^p) \leq k^{\epsilon/2}$. The first term in the upper bound is a constant, while the second term converges. Hence, the upper bound converges, and thus our series is convergent.
Therefore, the series converges for every $p>1$ and diverges for any $0<p\leq 1$. 
A: You can use the integral test. We consider the integral 
$$\int_{1}^{\infty}\frac{\ln(1+x^p)}{x^p}dx= \lim _{x\rightarrow \infty }{\frac {\ln  \left( 1+{x}^{p} \right) }{
 \left( 1-p \right) {x}^{p-1}}}+{\frac {\ln  \left( 2 \right) }{-1+p}}+\frac{p}{p-1}
\int _{1}^{\infty }\!{\frac {1}{   \left( 1+{x}^{p
} \right) }}{dx}.$$
Now, you can see that, the above limit diverges if $0<p<1$ together with the last integral, since
$$ \int _{1}^{\infty }\!{\frac {1}{   \left( 1+{x}^{p
} \right) }}{dx} \sim \int _{1}^{\infty }\!{\frac {1}{ {x}^{p
} }}{dx} $$
which makes the whole integral diverges and hence the series diverges. On the other hand, the limit is $0$ and the integral converges for $p>1$ which implies that the series converges.
A: We have
$$\frac{\ln\left(1+n^p\right)}{n^p}\sim_\infty\frac{\ln\left(n^p\right)}{n^p}=p\frac{\ln(n)}{n^p}$$
So


*

*if $0\leq p\leq 1$ then $\frac{\ln(n)}{n^p}\geq \frac{1}{n^p}$ and the given series is divergent.

*if $p>1$ then pick $p'$ such that $1<p'<p$ and we have $\frac{\ln(n)}{n^p}=_\infty o\left(\frac{1}{n^{p'}}\right)$ and the given series is convergent.

A: Try 
$$p\ln(n)=\ln(n^p)<\ln(1+n^p) < \ln(2n^p) =\ln(2) +p\ln(n)<2p\ln(n) $$
