# Asymptotic expansion of the sequence $u_n=\sum\limits_{k=1}^{n}e^{-b^{-k}}$

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 ?

## 3 Answers

Since $$e^{-x}=1-x+x^2/2!-x^3/3!+\ldots$$, your sum is \begin{align} \sum_{k=1}^n e^{-b^{-k}}&= n-\sum_{m=1}^\infty \frac{(-1)^{m+1}}{m!}\frac{1-b^{-mn}}{b^m-1} \\&=n-\frac{1-b^{-n}}{b-1}+\frac{1}{2}\frac{1-b^{-2n}}{b^2-1}-\frac{1}{6}\frac{1-b^{-3n}}{b^3-1}\ldots+\frac{(-1)^{m+1}}{m!}\frac{1-b^{-mn}}{b^m-1}+\ldots \end{align} Immediately, this tells us that $$u_n\sim n-\sum_{m=1}^\infty \frac{(-1)^{m+1}}{m!}\frac{1}{b^m-1}$$. If $$b$$ is large enough, then $$u_n\approx_{n\to\infty} n+e^{-b^{-1}}-1\sim n -\frac{1}{b}$$.

• The expansion for $1 - b^{-m n}$ cannot be correct ($b^{-m n}$ decays exponentially). The next terms in the expansion of $u_n$ after $n - \alpha$ are simply $$\sum_{m \geq 1} \frac {(-1)^{m - 1} b^{-m n}} {m! (b^m - 1)}.$$ – Maxim Feb 2 at 18:24
• Thank you, good job finding the third term of the expansion ! – Harmonic Sun Feb 2 at 19:47

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 $$b$$.

• Thank you very much ! – Harmonic Sun Feb 2 at 17:46

We have \begin{align*} u_n &= \sum_{k=1}^{n} \sum_{r=0}^\infty \frac{(-b^{-k})^r}{r!} \\ &= \sum_{r=0}^\infty \sum_{k=1}^{n} \frac{(-b^{-k})^r}{r!} \\ &= \sum_{r=0}^\infty \frac{(-1)^{r+1}}{r!}\frac{b^{-nr}-1}{b^r-1} \\ &= n + \frac{b^{-n}-1}{b-1} + \cdots \end{align*}

• This gives a more accurate expansion than mine, but for an asymptotic expansion can we not remove the $b^{-n}$ terms as they will become zero? – Peter Foreman Feb 2 at 17:28
• You're absolutely right, and your answer probably deserves more credit, because I answered effectively the same way, although maybe a few seconds too late haha. I wanted to be as general as possible, for then the OP can determine however s/he wanted to handle the defect terms $b^{-nr}$. – Tom Chen Feb 2 at 17:33