Convergence and Irrationality of $\frac{H_{(n,-n)}}{(n+1)^n}$ as $n$ approaches infinity

Question

We define $H_{(a,b)}$ as the $a^{th}$ harmonic number of class $b$. In other words, $$H_{(a,b)}=\sum_{k=1}^a \frac{1}{k^b}$$ More information about generalized harmonic numbers can be found here. Let us choose some positive integer $n$ and look at $H_{n,-n}$. We obtain $$H_{(n,-n)}=\sum_{k=1}^n {k^n}=1^n+2^n+3^n+...+(n-2)^n+(n-1)^n+n^n$$ We then define an error term $\tau$ such that $$H_{(n,-n)}=\tau (n+1)^n$$ $\tau$ is therefore the ratio between the sum of all integers up to $n$ raised to the $n$ and $n+1$ raised to the $n$. We can easily solve for $\tau$ and obtain $$\tau=\frac{H_{(n,-n)}}{(n+1)^n}$$ However, $\tau$ did not grow as I expected when $n$ approached infinity. It seemed to converge around $0.58196916011792172584$ (this is the approximate value of $\tau$ at $n=10000$). My calculations suggest that $\tau$ approaches a constant, but I do not like to rely on numerical evidence. Is it possible to prove that the value of $\tau$ converges as n approaches infinity? If so, what can be said about the constant it converges towards? Is it irrational?

Answer 1

$$\eqalign{\dfrac{H_{(n,-n)}}{(n+1)^n} &= \sum_{j=0}^{n-1} \left(1 -\dfrac{j+1}{n+1}\right)^n\cr &\approx \sum_{j=0}^{\infty} \exp(- (j+1))\cr &= \dfrac{1}{e-1}}$$