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

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.

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.

• Is it the case that $a_n > 2$ will always be true solely because we are assuming $a_n \rightarrow +\infty$? – dc3rd Oct 14 '18 at 20:08
• @dc3rd yes, past a certain point $N$, pick $M=2$ in the definition you quoted – Calvin Khor Oct 14 '18 at 20:26

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.

• 3. if I am going to pick an M, what comes to my mind is that based on the Archimedian property of $\mathbb{R}$ there iwll exist an M s.t $\frac{1}{M} < \epsilon$ So choose M to be large enough to satisfy this? 4. & 5. to relate $p$ to $k$, perhaps the statement that for all $k > p$? 2. Should I state for M > 0, to remove that possibility? – dc3rd Oct 14 '18 at 20:04
• Well $M$ doesn't actually have to be very large, you don't even need the tail sum to be small, just finite. – Calvin Khor Oct 14 '18 at 20:08
• What about my assumptions for 4. & 5.? – dc3rd Oct 14 '18 at 20:11
• I'm not sure what you mean there? – Calvin Khor Oct 14 '18 at 20:17
• @dc3rd all $k,p$ such that? You don't need something that complicated, you just need $k$ large so that $a_n>M=1$ say, then for $p>1$ small enough the inequality you are using holds. But you do not need the comparison to a $p$-series if this is too complicated, see the other answer – Calvin Khor Oct 14 '18 at 20:24