Check that $\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}\left(\frac{i+x}{n}\right)^n=\frac{e^{x+1}}{e-1}$

Question

Show that $$\lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{i+x}{n}\right)^n=\frac{e^{x+1}}{e-1}$$ Any hints how I can tackle this problem?

Although I checked on a sum calculator that it converges very slowly, this result gives me a reason that the proposed closed form is incorrect. Can anyone verify it?

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I know that $$\lim_{n\to\infty}\left(\frac{n+x}{n}\right)^n=e^x$$

The limit to be computed is

$$\lim_{n\to\infty}\left(\frac{1+x}{n}\right)^n+\left(\frac{2+x}{n}\right)^n+\left(\frac{3+x}{n}\right)^n+\cdots$$

It looks like the natural number series

$$1^n+2^n+3^n+4^n+\cdots$$

But this is the farthest I can go.

Answer 1

This solution uses the following facts:

• For every real number $t$, $\left(1+\frac{t}n\right)^n\to e^t$ when $n\to\infty$.
• For every real number $t$, $1+t\leqslant e^t$.
• For every real number $t\geqslant-\frac12$, $1+t\geqslant e^{t-t^2}$.

One is asking to prove that the limit of $$S_n(x)=\sum_{i=1}^{n}\left(\frac{i+x}n\right)^n=\sum_{k=0}^{n-1}\left(1+\frac{x-k}n\right)^n$$ when $n\to\infty$, exists and equals $$s(x)=\sum_{k=0}^{\infty}e^{x-k}=\frac{e^{x+1}}{e-1}.$$ To prove this, first note that $$S_n(x)=S_n(x-1)+\left(1+\frac{x}n\right)^n-\left(\frac{x}n\right)^n,$$ hence, for every $x$, $$\lim_{n\to\infty}\ (S_n(x)-S_n(x-1))=e^x.$$ Next, assume that $x\geqslant0$. Then, the bound $1+t\leqslant e^t$, valid for every $t$, and the fact that $1+\frac{x-k}n\geqslant0$ for every $k$ in the second sum above defining $S_n(x)$, yield $$S_n(x)\leqslant\sum_{k=0}^{n-1}\left(e^{(x-k)/n}\right)^n=\sum_{k=0}^{n-1}e^{x-k}\leqslant s(x).$$ Likewise, pick some $a$ in $(0,1)$ and assume that $n$ is large enough for $n^{1-a}\geqslant2$ to hold. Then, the bound $1+t\geqslant e^{t-t^2}$, valid for every $t\geqslant-\frac12$, and the fact that $1+\frac{x-k}n\geqslant0$ and that $\frac{x-k}n\geqslant-\frac12$ for every $k\leqslant n^a$, together yield $$S_n(x)\geqslant\sum_{k=0}^{n^a}\left(e^{(x-k)/n-(x-k)^2/n^2}\right)^n=\sum_{k=0}^{n^a}e^{x-k-(x-k)^2/n}\geqslant e^{-n^{2a-1}}\sum_{k=0}^{n^a}e^{x-k}=e^{-n^{2a-1}}s(x)\left(1-e^{-n^a}\right).$$ If $a$ is in $(0,\frac12)$, $e^{-n^{2a-1}}\to1$ and $e^{-n^a}\to0$, hence $S_n(x)\to s(x)$, thus the claim holds for every $x\geqslant0$.

Finally, the claim holds for every $x$ because $$s(x)-s(x-1)=e^x=\lim_{n\to\infty}\ (S_n(x)-S_n(x-1)).$$

Answer 2

We will take Taylor series of $$f_n(x) = \sum_{i=1}^n \left( \frac{i+x}{n} \right)^n$$

We are aiming for $\frac{e^{x+1}}{e-1} = A (1+x+\frac{x^2}{2} + \dots)$ where $A = \frac{e}{e-1}$.

The $x^0$ coefficient of the sum is $$f_n(0) = \sum_{i=1}^n \left(\frac{i}{n} \right)^n$$

If we differentiate the sum, we obtain $$f_n'(x) = \sum_{i=1}^n \left( \frac{i+x}{n} \right)^{n-1}$$ and in general if we differentiate $k$ times we obtain $$f_n^{(k)}(x) = \sum_{i=1}^n \left( \frac{i+x}{n} \right)^{n-k} \frac{(n-1) \dots (n-k+1)}{n^{k-1}}$$ Therefore $$\frac{f_n^{(k)}(0)}{k!} = \sum_{i=1}^n \left( \frac{i}{n} \right)^{n-k} \frac{1}{n^k} \binom{n}{k}$$

We therefore obtain the Taylor series of $f_n$; I claim that for $k$ fixed, $\frac{f_n^{(k)}(0)}{k!}$ tends to a limit as $n \to \infty$. From this limit, we can obtain the Taylor series of $\lim_{n \to \infty} f_n(x)$, and verify that it is equal to that of $\frac{e^{x+1}}{e-1}$.

From this point on is a sketch, because I really need to go to bed. (Strictly speaking, as well as filling out the details below, we also need to check various convergences; it's not in general true that having zero Taylor series means the function is everywhere zero, for instance.)

The limit of $\binom{n}{k} \frac{1}{n^k}$ is $\frac{1}{k!}$; the limit of $\sum_{i=1}^n \left( \frac{i}{n}\right)^{n-k}$ is the same for any $k$, and is equal to $\frac{e}{e-1}$.