Check convergence of given series

Question

$a_n≥0$ for all $n$ such that $a_n \to {\infty}$ as $n \to {\infty}$. Then there exist a natural number $M$ such that $\sum_{n=0}^{\infty}$$\frac{1}{(a_n) ^M}$ is convergent. (T/F)

My solution: There exist $K$ such that $a_n>1$ for all $n≥K$. Choose smallest natural number $m_n$ such that $(a_n)^{m_n}>n^2$ for all $n≥k$.

If ${m_n}$ has sup $M$, then $M$ works as our choice. Now, since $a_n$ diverges to $\infty$, after finitely many terms ${m_n}$ becomes bounded. But I can't show it Rigorously.

Any another strategy is also appreciable.

Answer 1

It is false.

Let $a_n=\ln n$,

By integral test, let $t=\ln x$

$$\int_2^\infty \frac{1}{\ln^M x} dx=\int_{\ln2}^\infty t^{-M}e^t dt\longrightarrow \infty$$

Answer 2

Another counter example: $a_n = \log n$. For any $M > 0$ you can always choose an $0 < \alpha < 1$, and a natural number $K$ such that $(\log n)^M < n^{\alpha}, n \ge K$. Then $\displaystyle \sum_{n=K}^\infty \dfrac{1}{(\log n)^M}> \displaystyle \sum_{n=K}^\infty \dfrac{1}{n^{\alpha}} = \infty$