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Evaluate $$ \lim_{n\to+\infty}\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1). $$

Since $\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1)\leq n(e^{\frac{1}{n}})-n$, which implies that the limit is no more than $1$. But I met some problems in the method of enlarging and reducing.

Knt
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  • But $\lim n(e^{1/n}-1)$ is $e$, not $1$ – Hagen von Eitzen Oct 12 '19 at 09:12
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    HINT: Try proving $x + x^2 \ge e^x-1 \ge x$ whenever say $0<x< \frac{1}{N}$ for some big $N$. Then notice for $n>N$ we have $\frac{k^2}{n^3} < \frac{1}{N}$ for any $k \in {0,...,n}$. Use first inequality (why we can?), Riemann sums and squezee theorem – Presage Oct 12 '19 at 09:15

3 Answers3

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As you refer to reducing and enlarging, squeezing is a clean way here together with Riemann sums:

To do so, apply Taylor to $e^x$ with 2nd degree remainder and bound the remainder:

  • $x> 0 \Rightarrow e^x-1 = x+ \frac{e^{\xi}}{2}x^2 \mbox{ with } 0<\xi <x$
  • $\stackrel{0 \leq x\leq 1}{\Rightarrow} x + \frac{1}{2}x^2 \leq e^x-1 \leq x + \frac{e}{2}x^2$

Applying this to your sum you get

$$\underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^2\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^2\;dx} + \frac{1}{2n}\underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^4\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^4\;dx} \leq \sum_{k=1}^n\left(e^{\frac{k^2}{n^3}}-1\right) \leq \underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^2\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^2\;dx} + \frac{e}{2n}\underbrace{\sum_{k=1}^n\left(\frac{k}{n}\right)^4\frac{1}{n}}_{\stackrel{n\to \infty}{\longrightarrow}\int_0^1x^4\;dx}$$

Now, sending $n\to \infty$ gives $\lim_{n\to \infty}\sum_{k=1}^n\left(e^{\frac{k^2}{n^3}}-1\right) = \int_0^1x^2\;dx = \frac{1}{3}$.

trancelocation
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We have that

$$e^{\frac{k^2}{n^3}}-1=\frac{k^2}{n^3}+O\left(\frac{k^4}{n^6}\right)$$

and therefore

$$\sum_{k=1}^{n}(e^{\frac{k^2}{n^3}}-1)=\frac1{n^3}\sum_{k=1}^{n}k^2+\frac1{n^6}\sum_{k=1}^{n}O(k^4)$$

then refer to Faulhaber's formula

user
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$$L=\lim_{n \rightarrow \infty}\sum_{k=1}^{n} (e^{k^2/n^3}-1)= \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \left( [(k^2/n^2)+ \frac{1}{2}(k^4/n^5)+... ]\right).$$ Let $k/n=x$, then $k^4/n^5 \rightarrow 0$, when $n\rightarrow \infty.$ Then $$L=\int_{0}^{1} x^2 dx=\frac{1}{3}.$$

Z Ahmed
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