calculus and series If we suppose $\lim_{n\to+\infty} a_n =0,$ I have to examine if $\sum _{ n=1 }^{ \infty  } n^{\frac {1}{1+a_n}}$ converges or diverges.
I used the monotone convergence theorem to show that the series diverges. If we suppose: $a_n=\dfrac {1}{n},$ then $$\lim_{n\to+\infty} a_n= \lim_{n\to+\infty} \frac {1}{n}=0. $$ Let the sequence of functions $f$ on $\mathbb{N}$ be defined by $$f_n (k)= k^{\frac{1}{1+\frac {1}{n}}}.$$ We note that these functions are positive, and that for each $k$, as a sequence in $n, f_n (k)$ is increasing. The sum over $k$ is an integral and it's the integral over $\mathbb{N}$ with respect to counting measure. In particular, by the Monotone Convergence Theorem (from the Lebesgue theory) allows to interchange limit and integral:
$$\lim_{n\to+\infty} \sum _{ k=1 }^{ \infty  } k^{\frac {1}{1+\frac {1}{n}}} = \sum _{ k=1 }^{ \infty  } \lim_{n\to+\infty} k^{\frac {1}{1+\frac {1}{n}}}= \sum _{ k=1 }^{ \infty  } \frac{1}{k}= +\infty$$
and then $$ \sum _{ k=1 }^{ \infty  } k^{\frac {1}{1+\frac {1}{n}}} $$ diverges. By the same argument, $$\sum _{ k=1 }^{ \infty  } k^{\frac {1}{1+a_n}}$$  diverges.

Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n \geq 1$ and $x \geq 0$.
