The value of $\sum_{n=1}^{\infty}\frac{1}{n^n}$?

Question

In my last question I asked about riemann zeta function $\zeta(3)$ and got an answer as apery's constant which is an unsimplifiable transcendental constant. I don't know why but I just got more curious and I want to find out $$\sum_{n=1}^{\infty}\frac{1}{n^n}$$As it comes to $1/1, 1/4, 1/27, 1/128$ and so on I think it'll be somewhere close to $1.3$. However, it's just a guess from the first few values and is very inaccurate. I'll be delighted to hear some answers.

Answer 1

$\sum_{n=1}^{\infty}\frac{1}{n^n}$ is well known as Sophomore's Dream.

For the proof, you can refer here

https://en.wikipedia.org/wiki/Sophomore%27s_dream

(wiki contains a relatively elementary proof)

Answer 2

Just a few accurate numbers for your curiosity. $$S_p=\sum_{n=1}^{p}\frac{1}{n^n}$$ $$\left( \begin{array}{cc} p & S_p \\ 1 & 1.0000000000000000000000000000000000000000000000000 \\ 2 & 1.2500000000000000000000000000000000000000000000000 \\ 3 & 1.2870370370370370370370370370370370370370370370370 \\ 4 & 1.2909432870370370370370370370370370370370370370370 \\ 5 & 1.2912632870370370370370370370370370370370370370370 \\ 6 & 1.2912847205075445816186556927297668038408779149520 \\ 7 & 1.2912859347732234836310582023740719585201114218848 \\ 8 & 1.2912859943778682590216832023740719585201114218848 \\ 9 & 1.2912859969590430507348803843641034666368646377944 \\ 10 & 1.2912859970590430507348803843641034666368646377944 \\ 11 & 1.2912859970625479897296943093401434997684906233581 \\ 12 & 1.2912859970626601463844789244244143783796628889530 \\ 13 & 1.2912859970626634480754341545436790897428183570720 \\ 14 & 1.2912859970626635380681794523564896249717405834021 \\ 15 & 1.2912859970626635403518377128776568469768730997672 \\ 16 & 1.2912859970626635404060478215019320686772457398107 \\ 17 & 1.2912859970626635404072566601502344653905647439530 \\ 18 & 1.2912859970626635404072820754307878088462831218953 \\ 19 & 1.2912859970626635404072825808847227912309471354275 \\ 20 & 1.2912859970626635404072825904214659552934471354275 \\ 21 & 1.2912859970626635404072825905926230124075490460176 \\ 22 & 1.2912859970626635404072825905955518886543273546023 \\ 23 & 1.2912859970626635404072825905955997803012007524468 \\ 24 & 1.2912859970626635404072825905956005300748945807507 \\ 25 & 1.2912859970626635404072825905956005413338936491770 \\ 26 & 1.2912859970626635404072825905956005414963336386819 \\ 27 & 1.2912859970626635404072825905956005414985888039157 \\ 28 & 1.2912859970626635404072825905956005414986189739026 \\ 29 & 1.2912859970626635404072825905956005414986193633583 \\ 30 & 1.2912859970626635404072825905956005414986193682152 \\ 31 & 1.2912859970626635404072825905956005414986193682738 \\ 32 & 1.2912859970626635404072825905956005414986193682745 \\ 33 & 1.2912859970626635404072825905956005414986193682745 \end{array} \right)$$ and if you want $$S_{p}-S_{p-1} \leq 10^{-k}$$ you need to solve $$p^p=10^k \implies p=\left\lceil \frac{k \log (10)}{W(k \log (10))}\right\rceil$$ where $W(.)$ is Lambert function.

For $k=50$, this would give $p=33$ as the table shows.