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Find $$\lim_{n\rightarrow \infty }\left ( \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n}+.....+ \frac{e^{\frac{n-1}{n}}}{n}\right ).$$

Solving a bit and applying GP, I got

$\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{n.\left ( e^{\frac{1}{n}} -1 \right )}$

Now, limit gives the expression as

$\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{\infty *0}$

How do I find it now? Should I use the $\frac{0}{0}$ form?

  • 1
    $$\dfrac{e^\frac1n-1}{\frac1n}=\frac{\frac1n+o(\frac1n)}{\frac1n}\to 1.$$ Or still another approach using Riemann integral: $$\lim_{n\to\infty}\sum_{i=0}^{n-1}\dfrac{e^\frac{i}n}{n}=\int_0^1 e^x\,\mathrm dx$$ – 2017-01-30
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    Hint: Riemann sums. – 2017-01-30
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    Notation police has arived: $\lim_{x\rightarrow \infty }\left ( \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n}+.....+ \frac{e^{\frac{n-1}{n}}}{n}\right )=\left ( \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} + \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n}+.....+ \frac{e^{\frac{n-1}{n}}}{n}\right )$, there is no $x$ in the expression. – 2017-01-30
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    Generally speaking, if you have an idea, you should *try* it. In all but the simplest of problems, you won't know what steps lead to a solution until you actually try taking the steps and find one that leads to a solution. – 2017-01-30
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    Your working is unclear: you have an expression as what you got after GP (which is undefined) and that doesn't tell us what what you got. – 2017-01-30

4 Answers 4

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You can continue with your way of thinking. Using the geometric progression formula we have that the sum is equal to:

$$\lim_{n \to \infty} \frac{e- 1}{n(e^{\frac 1n}-1)}$$

Now you can switch to real numbers and use the fact that:

$$\lim_{n \to \infty} \frac{1}{n(e^{\frac 1n}-1)} = \lim_{x \to \infty} \frac{1}{\frac{e^{\frac 1x}-1}{\frac 1x}} = \frac{1}{\lim_{x \to \infty} \frac{e^{\frac 1x}-1}{\frac 1x - 0}} = \frac{1}{(e^x)'|_{x=0}} = \frac{1}{e^0} = 1$$

Hence the sum is equal $e-1$

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    Ah, beautifully done. – 2017-01-30
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Hint: this is the Riemann sum of the following integral:

$$\int_0^1e^x\ dx$$

Which is easily solved to give $e-1$.

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    @JonGarrick No, it's indeterminate form. – 2017-01-30
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You were already pretty close with your own way!:

$$\lim_{n\to\infty}\frac1{n\left(e^{1/n}-1\right)}\stackrel{x:=\frac1n}=\lim_{x\to0}\frac x{e^x-1}$$

And now observe that

$$\lim_{x\to0}\frac{e^x-1}x=\left(e^x\right)'|_{x=0}=e^0=1$$

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@Jon Garrick: just forwarding your answer $$\displaystyle\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{n.\left ( e^{\frac{1}{n}} -1 \right )}$$ $$\displaystyle\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{n.\left ( 1+\frac1n+O(\frac1{n^2}) -1 \right )}=\displaystyle\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{\left ( 1+O(\frac1{n})\right )}=e-1$$