Three Same Variables in a Typical Element

Note that for $N\geq 2$ , $0 \leq \sum_{n=N+1}^\infty \frac{n}{n^n} \leq \sum_{n=N+1}^\infty \frac{1}{N^{n-1}} = \frac{1}{N^{N-1}(N-1)} \leq \frac{2}{N^N}$ so that in particular, for $N \geq 2$ , $\sum_{n=1}^N \frac{n}{n^n} \leq \sum_{n=1}^\infty \frac{n}{n^n} \leq \frac{2}{N^N} + \sum_{n=1}^N \frac{n}{n^n}$ which means our estimate of the series can be $\sum_{n=1}^N \frac{n}{n^n} + \frac{1}{N^N}$ which has an error of at most $\frac{1}{N^N}$ . Therefore, to be sure about our estimate, we would need $N$ to at least satisfy $\frac{1}{N^N} < \frac{1}{100000} \iff N \geq 7$

Therefore, for $N=7,$ we compute $\sum_{n=1}^7 \frac{n}{n^n} = 1.62847321 \\ \sum_{n=1}^7 \frac{n}{n^n} + \frac{2}{7^7} = 1.62847564$ which unfortunately isn't precise enough to know how it rounds to the requested precision, so we do one more term, with $N=8,$ to compute $\sum_{n=1}^8 \frac{n}{n^n} = 1.62847369 \\ \sum_{n=1}^8 \frac{n}{n^n} + \frac{2}{8^8} = 1.62847381$ which works!

That is, we know that $1.62847369 \leq \sum_{n=1}^\infty \frac{n}{n^n} \leq 1.62847381$ so that in particular, it rounds to the nearest millionth as $\sum_{n=1}^\infty \frac{n}{n^n} \approx \boxed{1.628474}$