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.


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.

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

  • $\begingroup$ 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? $\endgroup$ – dc3rd Oct 14 '18 at 20:04
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    $\begingroup$ Well $M$ doesn't actually have to be very large, you don't even need the tail sum to be small, just finite. $\endgroup$ – Calvin Khor Oct 14 '18 at 20:08
  • $\begingroup$ What about my assumptions for 4. & 5.? $\endgroup$ – dc3rd Oct 14 '18 at 20:11
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    $\begingroup$ I'm not sure what you mean there? $\endgroup$ – Calvin Khor Oct 14 '18 at 20:17
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    $\begingroup$ @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 $\endgroup$ – Calvin Khor Oct 14 '18 at 20:24

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