Determine whether this is true: If $a_n \rightarrow +\infty$ then $\sum_{n = 1}^{\infty}\frac{1}{a_{n}^{n}}$ converges

Question

Determine whether the following is true: If sequence $a_n \rightarrow +\infty$ then the infinite series $\sum_{n = 1}^{\infty} \frac{1}{a_{n}^{n}} $ converges.

Note: I am looking for feedback on whether my solution is correct or if I am at least on the right track.

Attempt:

A sequence $a_n \rightarrow +\infty$ means:

$$ \forall\ M\in \mathbb{R} \ \exists \ N\in \mathbb{N}\ \ s.t \ \forall\ n \geq N,\ a_n > M$$

$$ \Leftrightarrow \forall\ M\in \mathbb{R} \ \exists \ N\in \mathbb{N}\ \ s.t \ \forall\ n \geq N,\ \frac{1}{M} > \frac{1}{a_n}$$

So let $\epsilon > 0$, This means we have:

$$\Bigg| \sum_{n = k+1}^{\infty} \frac{1}{a_{n}^{n}} \Bigg| \leq \sum_{n = k+1}^{\infty} \Bigg| \frac{1}{a_{n}^{n}} \Bigg| < \sum_{n = k+1}^{\infty} \Bigg| \frac{1}{M^{n}} \Bigg| < \sum_{n = k+1}^{\infty} \Bigg| \frac{1}{n^{p}} \Bigg| < \epsilon $$

It has been shown that $$\sum_{n = k+1}^{\infty} \Bigg| \frac{1}{n^{p}} \Bigg|$$ converges for $p > 1$ now since $n \in \mathbb{N},\ n > 1$ eventually. So the original series converges.

Answer 1

You are bringing in the $p$-test unnecessarily. Once you know that all the terms in the tail of the sequence satisfy $a_n > 2$, say when $n>N$, then you can compare $$ \bigg|\sum_{n > N}\frac{1}{a_n^n}\bigg| \le \underbrace{\sum_{n> N}\frac{1}{2^n}}_{\text{geometric series}} = \frac{1}{2^N}, $$ hence the original series converges.

Answer 2

It seems that you got the idea right. But some remarks on your solution -

1. You switched between $a_n$ and $x_n$
2. the "iff" is not correct, because the second line is solved by any $x_n < 0$ as well
3. $M$ in the "iff" was quantified over; you should pick one before making the lines in the computation
4. $k$ appears suddenly; does it matter what $k$ is?
5. $p$ appears suddenly; perhaps you want it to relate to $k$?

Lets attempt to fix this, as if we forgot that the geometric series converged...

First lets choose some $M>1$. then $a_n >M$ when $n>k$ for some $k$. Then, note that $M^k > k^p$ is the same as $p < \frac{k}{\log k}\log M$, so if necessary make $k$ bigger so that $p=2$ is allowed. Then for $n>k$, $M^n > n^p$ $$ \sum_{n=k+1}^K\frac1{M^k}\le\sum_{n=k+1}^K \frac1{n^2} < \pi^2/6$$ so the tail sum converges.