Convergence of $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^k}$ through Cauchy criterion

Question

So I need to show that the series $$ \sum_{k=1}^{\infty}\frac{1}{k^k}$$ converges, which is very quick using different tests, but the caveat is that I need to do this using the Cauchy criterion for series. This boils down to showing the sequence of partial sums $(s_n)$ is Cauchy, but the problem lies in actually finding a closed form for $(s_n)$.

I tried checking by hand the cases $n=1, 2, 3, 4, 5$ to find a pattern, make an educated guess and prove it correct by induction, but no clear pattern arises.

There is no closed form. See also art of problem solving, or this duplicate. — Dietrich Burde, Mar 12 '24 at 17:16
So... you don't need to find a closed form for the partial sums themselves, just estimates that satisfy the Cauchy criterion. — Sammy Black, Mar 12 '24 at 17:17
Does this answer your question? "Closed" form for $\sum \frac{1}{n^n}$ — Dietrich Burde, Mar 12 '24 at 17:20
For $k\ge 2$, the series converges faster than the series $\sum_{k\ge 2}\frac1{k^2}$. — Mark Viola, Mar 12 '24 at 17:20

score 2 · Answer 1 · answered Mar 12 '24 at 17:59

According to the Cauchy criterion applied to the partial sums $(s_n)$, given $\varepsilon > 0$, we need to find some $N \in \mathbb{N}$ such that whenever $m \geq n \geq N$, $$ |s_m - s_n| = \Biggl|\, \sum_{n < k \leq m}\!\! a_k \,\Biggr| < \varepsilon. $$

As long as you can find some other series $(b_n)$ such that $$ 0 < a_n \leq b_n $$ and the series $\displaystyle\sum_{1 \leq n} b_n$ is Cauchy, you're good. Try to find such a series before revealing the spoiler.

In your series, the summands are $a_n = n^{-n} \leq 2^{-n}$ for any $n \geq 2$, and $$\begin{align} \sum_{n < k \leq m}\!\! k^{-k} &\leq \sum_{n < k \leq m}\!\! 2^{-k} \\ &= \sum_{0 < k \leq m}\!\! 2^{-k} - \sum_{0 < k \leq n}\!\! 2^{-k} \\ &= \frac12 \cdot \frac{1 - 2^{-m}}{1 - 2^{-1}} - \frac12 \cdot \frac{1 - 2^{-n}}{1 - 2^{-1}} \\ &= \bigl( 1 - 2^{-m} \bigr) - \bigl( 1 - 2^{-n} \bigr) \\ &= 2^{-n} - 2^{-m} \\ &\leq 2^{-n}, \end{align}$$ so choosing $N > -\frac{\ln\varepsilon}{\ln 2}$ suffices to yield $2^{-n} \leq 2^{-N} < \varepsilon$.

I originally tried this using a geometric series, but the details became too laborious for me. Nicely done! — K.defaoite, Mar 12 '24 at 18:05

score 1 · Answer 2 · answered Mar 12 '24 at 18:03

Consider $s(n)=\sum_{k=1}^nk^{-k}$. Cauchy's criterion requires that we show that , for all $\epsilon>0$, there exists a $N\in\Bbb N$ such that

$$|s(n+m)-s(n)|\leq \epsilon \\ m\in\mathbb N$$ For all $n> N$.

In our case, $$|s(n+m)-s(n)|=\sum_{k=n+1}^{n+m}k^{-k}=\sum_{k=1}^m (n+k)^{-(n+k)}$$ We know, for certain, that $$|s(n+m)-s(n)|\leq \sum_{k=1}^m (n+k)^{-2}$$ By simply comparing terms.

We can bound this sum with an integral (draw a graph!) $$\sum_{k=1}^m (n+k)^{-2}\leq \int_1^m \frac{1}{(n+x-1)^2}\mathrm dx=\frac{1}{n}-\frac{1}{n+m-1}\leq \frac{1}{n}$$ Therefore $$|s(n+m)-s(n)|\leq \frac{1}{n}$$

Therefore, for any given $\epsilon$, we can simply choose an $N$ such that $\frac{1}{N}<\epsilon$. $\blacksquare$

Convergence of $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^k}$ through Cauchy criterion

2 Answers2