Estimating an exponential sum $\sum_{n = 1}^x e^n/n$

Question

I'm interested in estimating the exponential sum $$ \sum_{n = 1}^x \frac{e^n}{n} $$ for reasonably large $x$.

One way to try to understand this sum is to approximate it with an integral $$ \sum_{n = 1}^x \frac{e^n}{n} \leq \frac{e^x}{x} + \int_1^x \frac{e^t}{t} dt, $$ which would lead us to estimate the integral version. The integral is closely related to the classical exponential integral $\text{Ei}(x)$, except that we only pay attention to the worst-converging part.

For either the integral or the sum, the summand/integrand $e^t/t$ grows so quickly that the entire sum/integral should be dominated by the last unit interval. In particular, the sum looks almost geometric with ratio $e$. I haven't made this rigorous, but this sort of reasoning suggests that $$ \sum_{n = 1}^x \frac{e^n}{n} \leq 2 \frac{e^x}{x}. $$

How can we estimate that sum, either by improving one of these arguments or otherwise?

Answer 1

Asymptotic series for sums like this where the largest contributions come from only a few terms of the sum (in this case the terms near $n=x$) can be found by expanding the subdominant factors in power series in this region.

We'll assume that $x$ is an integer, which allows us to set $m=x-n$ and rewrite the sum like

$$ \sum_{n=1}^{x} \frac{e^n}{n} = e^x \sum_{m=0}^{x-1} \frac{e^{-m}}{x-m}. $$

Now the largest contribution to the sum comes from a neighborhood of $m=0$, and we expand

$$ \frac{1}{x-m} = \frac{1}{x} \sum_{j=0}^{\infty} \left(\frac{m}{x}\right)^j. $$

We can substitute this back into the sum to get the asymptotic series

$$ \sum_{n=1}^{x} \frac{e^n}{n} \sim x^{-1}e^x \sum_{j=0}^{\infty} x^{-j} \sum_{m=0}^{x-1} m^j e^{-m}. $$

By attaching the tails to the inner sums we introduce errors which are exponentially smaller than the terms of the series, and thus obtain

$$ \sum_{n=1}^{x} \frac{e^n}{n} \sim x^{-1} e^x \sum_{j=0}^{\infty} x^{-j} \sum_{m=0}^{\infty} m^j e^{-m}. \tag{$*$} $$

We calculate

$$ \sum_{m=0}^{\infty} m^j e^{-m} = \begin{cases} \frac{e}{e-1} & j=0, \\ \frac{e}{(e-1)^2} & j=1, \\ \frac{e(e+1)}{(e-1)^3} & j=2, \end{cases} $$

so the first few terms of the asymptotic series for your sum are

$$ \sum_{n=1}^{x} \frac{e^n}{n} = \frac{e}{e-1} x^{-1} e^x + \frac{e}{(e-1)^2} x^{-2} e^x + \frac{e(e+1)}{(e-1)^3} x^{-3} e^x + O\!\left(x^{-4} e^x\right). $$

These calculations can be justified rigorously in a routine manner, e.g. as in §3.2 of de Bruijn's Asymptotic Methods in Analysis.

Answer 2

I found that $\sum_{n=1}^N \frac{e^n}{n} < \frac{e^{N+1}}{N+1}$. Their values are of the same order. For example, for $N = 10^6$, the sum gives $4.8\times10^{434288}$ while the approximation gives $8.25\times10^{434288}$. In fact, I would go further to suggest that $\sum_{n=1}^N \frac{e^n}{n} \sim \frac{e^{N+1}}{N+1}$. Still not sure how to go about showing this, maybe induction will work.

Answer 3

I will here derive the asymptotic $$S(N) = \sum_{n=1}^N \frac{e^n}{n} \sim \frac{e}{e-1}\frac{e^N}{N}$$

---

Massaging the series we see that

$$S(N) = \frac{e^N}{N}\left[1 + \frac{1}{e(1-1/N)} + \frac{1}{e^2(1-2/N)} + \ldots \right]\\= \frac{e^N}{N}\left[\frac{e}{e-1} + \frac{1}{e(N-1)} + \frac{2}{e^2(N-2)} + \ldots + \frac{N-1}{e^{N-1}}\right]$$

where I have added $0 = \frac{e}{e-1} - 1 - \frac{1}{e} - \frac{1}{e^2}-\ldots$ to reach the last line. The ratio of two successive terms in the series above is

$$\frac{1+\frac{1}{k}}{e\left(1 - \frac{1}{n-k}\right)} \leq \frac{2}{e\left(1 - \frac{1}{n-1}\right)} \leq \frac{2+\epsilon}{e} < 1$$

for any $0 < \epsilon < e-2$ as long as $n$ is large enough. We can therefore bound the series above by a geometrical series giving us the upper bound

$$S(N) \leq \frac{e^N}{N}\left[\frac{e}{e-1} + \frac{1}{(N-1)} \frac{1}{e-2-\epsilon}\right] \implies \lim_{N\to\infty} \frac{NS(N)}{e^N} \leq \frac{e}{e-1}$$

To establish a lower bound take $g(z) = \sum_{n=1}^N \frac{x^n}{n}$ to get $g'(z) = \frac{x^N-1}{x-1}$ and by integrating up we get the integral identity

$$S(N) = H_N + \int_1^e \frac{z^N-1}{z-1}{\rm d}z$$

where $H_N = \sum_{n=1}^N\frac{1}{n} \sim \log(N)$. We can now use $\frac{z^N-1}{z-1} \geq \frac{z^N-1}{e-1}$ for $z\in[1,e]$ which in the integral identity above leads to

$$\lim_{N\to\infty} \frac{S(N) N}{e^N} \geq \frac{e}{e-1}$$

and the desired result follows.

Answer 4

One might note that

$$\begin{align}\sum_{n=1}^x\frac{e^n}n&=e\sum_{n=1}^x\int_0^et^{n-1}~\mathrm dt\\&=e\int_0^e\sum_{n=1}^xt^{n-1}~\mathrm dt\\&=e\int_0^e\frac{1-t^x}{1-t}~\mathrm dt \\&=eH_x+e\int_1^e\frac{1-t^x}{1-t}~\mathrm dt\end{align}$$

Where $H_x$ is the $x$th Harmonic number. Use your favorite approximation ;). Thus, the remaining task is estimating the rest of the integral:

For $x>2$,

$$\int_1^e\frac{1-t^x}{1-t}~\mathrm dt<ex+\frac12\left(\frac{e^x-1}{e-1}-x\right)^2$$

Which follows from trapezoidal rule and the fact that $\frac{1-t^x}{1-t}$ is convex with respect to $t$ for $x>2$.

A lower bound may be constructed using a tangent line:

$$\int_1^e\frac{1-t^x}{1-t}~\mathrm dt>ex+\frac{(e-1)^2}4x^2(x-1)^2$$