Let $\left(u_n\right)_{n\in\mathbb{N^*}}$ be the sequence defined $\forall n\in\mathbb{N^*}$ by :
$$u_n=\sum\limits_{k=1}^{n}e^{-b^{-k}}$$
With $b\in(1,+\infty)$.
We can instictively say that when $n\to\infty$, since $\lim\limits_{k\to\infty}e^{-b^{-k}}\to 1$, it will tend to be like $\sum\limits_{k=1}^{n}1=n$, and thus I expect the dominant term of the asymptotic expansion of $\left(u_n\right)$ to be $n$ :
$$u_n=n+o(n)$$
And numerical simulation shows indeed that $u_n\sim n$ for large n.
Now, this is no demonstration ; and not only am I trying to prove it rigorously, but also (and mainly) to expand it a bit more and find out the next few terms (well, at least the second term) of the asymptotic expansion…
Any suggestion ?
Best Answer
If we use the Taylor series expansion of $e^x$: $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ We have the following asymptotic estimation for your sequence (as for large $n$ we have $n \sim \infty$): $$u_n \sim \sum_{k=1}^{\infty}\sum_{n=0}^{\infty} \frac{(-b^{-k})^n}{n!}$$ $$=\lim_{n \to \infty}\sum_{k=1}^{n}\Big(1-\frac{1}{b^k}+\frac{1}{2! \cdot b^{2k}}-\frac{1}{3! \cdot b^{3k}}+...\Big)$$ $$= n-\frac{1}{b-1}+\frac{1}{2! \cdot (b^2-1)} - \frac{1}{3! \cdot (b^3-1)}+...$$ $$=n+\sum_{k=1}^{\infty}\frac{(-1)^k}{k! \cdot(b^k-1)}$$ In the above expansion I used the fact that $1+x+x^2+...=\frac{1}{1-x}$ which is valid as $|\frac{1}{b^k}| \lt 1$ for the given range of $b$.