How to prove that $\sum_{n=1}^{\infty}\frac{a_n}{1+n^\alpha a_n}$ diverges if $\sum_{n=1}^{\infty}{a_n}$ diverges and $\alpha < 1$?

Question

If $$\sum_{n=1}^{\infty}{a_n}$$ diverges and $$\alpha < 1$$ how can I prove that $$\sum_{n=1}^{\infty}\frac{a_n}{1+n^\alpha a_n}$$ diverges?

Answer 1

$\bullet$ Assuming that $\alpha >0$. Let $k$ be an integer number such that $k\alpha >1$. If we choose $a_{1}=a_{2^k}=a_{3^k}=\cdots=1$ and $a_n=0$ otherwise, then $\sum_{n=1}^\infty a_n=\infty$ and $$\sum_{n=1}^\infty \frac{a_n}{1+n^\alpha a_n}=\sum_{n=1}^\infty \frac{a_{n^k}}{1+(n^k)^\alpha a_{n^k}}=\sum_{n=1}^\infty \frac{1}{1+n^{k\alpha}}<\infty.$$

$\bullet$ In the cases that $\alpha < 0$: Let $a_n =n^{\lambda \alpha}$, $\lambda >0$, we have $$\sum_{n=1}^\infty \frac{a_n}{1+n^\alpha a_n}=\sum_{n=1}^\infty \frac{n^{\lambda \alpha}}{1+ n^{\alpha +\lambda \alpha}}=\sum_{n=1}^\infty \frac{1}{n^{-\lambda \alpha}+n^\alpha}\leqslant \sum_{n=1}^\infty \frac{1}{n^{-\lambda \alpha}}.$$ Choosing $\lambda >0$ such that $\lambda \alpha <-1$ then the last series converges.

$\bullet$ In the case $\alpha =0$. Assume that $\sum_{n=1}^\infty a_n =\infty$, we will prove that $\sum_{n=1}^\infty \frac{a_n}{1+a_n} =\infty$. If $(a_n)_n$ is bounded, i.e. there exists $C>0$ such that $a_n \leqslant C$ for all $n\geqslant 1$, then $$\frac{a_n}{1+a_n}\geqslant \frac{a_n}{C+1}$$ for all $n\geqslant 1$. Hence, $\sum_{n=1}^\infty \frac{a_n}{1+a_n} \geqslant \sum_{n=1}^\infty \frac{a_n}{C+1}=\infty.$ In the remaining case $(a_n)_n$ is unbounded, there exists a sub-sequence $(a_{n_k})_{k\geqslant 1}$ such that $a_{n_k}\geqslant k$ for all $k$, we obtain $$\sum_{n=1}^\infty \frac{a_n}{1+a_n}\geqslant \sum_{k=1}^\infty \frac{a_{n_k}}{1+a_{n_k}}\geqslant \sum_{k=1}^\infty \frac{k}{1+k}=\infty.$$