which of the following series convergent?

Question

which of the following series converges?

A. $\sum_\limits{n=1}^\infty \frac{1}{3^{\ln n}}$

B. $\sum_\limits{n=1}^\infty \frac{1}{2^{\ln n}}$

C. $\sum_\limits{n=1}^\infty \frac{1}{n^{1+ \frac{1}{n} } }$

D.$\sum_\limits{n=1}^\infty \frac{ \pi^n}{n+ e^{ n}}$

I think that D is divergent( the general term doesn't go to zero) $\lim_{n \to\infty} \frac{ \pi^n}{n+ e^{ n}} = \infty$ .

For C it is easy to see using pre-calculus tools that $\frac1{n^{1+1/n}}$ decreases monotonically for $n\ge 1$ with $$\lim_{n\to \infty}\frac 1{n^{1+1/n}}=\lim_{n\to \infty}\frac{1}{n^{1+\frac{1}{n}}}=\lim_{n\to \infty}\frac{1}{n^{\phantom{\frac{1}{n}}}}\frac{1}{n^{\frac{1}{n}}}=0$$ but i don't know that $\sum_\limits{n=1}^\infty \frac{1}{n^{1+ \frac{1}{n} } }$ is diverge or converge .for B we know that :

let $ {a_n }$ be non negative sequance and $ {a_n } \to 0$ and $ {a_{n+1} } \le {a_{n} } $ then $\sum_\limits{n=1}^\infty a_n$ is converge iff $\sum_\limits{n=1}^\infty 2^n a_{ 2^n}$ be converge

$\sum_\limits{n=1}^\infty \frac{1}{2^{\ln n}}$ is converge iff $\sum_\limits{n=1}^\infty 2^n \frac{1}{2^{\ln 2^n}} =\sum_\limits{n=1}^\infty 2^{n(1-\ln 2)}$ be converge but it is clear that $\sum_\limits{n=1}^\infty 2^n \frac{1}{2^{\ln 2^n}}$ is diverge and yet $\sum_\limits{n=1}^\infty \frac{1}{2^{\ln n}}$ is diverge.

Answer 1

Hint for A,B:

$$3^{\ln n} = n^{\ln 3},\,\,\,\,2^{\ln n} = n^{\ln 2}.$$

(You can verify these equalities by applying $\ln$ to both sides.)