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I am trying to understand this:

$\displaystyle \sum_{n=1}^{\infty} e^{-n}$ using integrals, what I have though:

$= \displaystyle \lim_{m\to\infty} \sum_{n=1}^{m} e^{-n}$

$= \displaystyle \lim_{m\to\infty} \frac{1}{m}\sum_{n=1}^{m} me^{-n}$

So, suppose this is an right-hand Riemann sum, with $m$ Equal subintervals.

$f(x_i) = me^{-n}$ represents the height of the function, we will have the integral for.

$\Delta(x) = \frac{1}{m}$

But, How can this be represented as an integral?

Thanks!

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    $\begingroup$ This sum can be computed exactly with a geometric series, $$\sum_{n=1}^\infty \left(\frac{1}{e}\right)^n = \frac{1/e}{1-(1/e)} = \frac{1}{e-1}$$ $\endgroup$ – Joel Oct 30 '14 at 14:13
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    $\begingroup$ @Joel: OP probably knows this. He is trying to use a different method to arrive at the same result. This is an exercise in Riemann sums. $\endgroup$ – MPW Oct 30 '14 at 14:26
  • $\begingroup$ @MPW, thanks, exactly, I am trying to explore this integral method. $\endgroup$ – Amad27 Oct 30 '14 at 14:29
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    $\begingroup$ It's not meant to be an answer. That's why it's a comment. $\endgroup$ – Joel Oct 30 '14 at 14:46
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Maybe it's not what you want, but it uses integral method...

We compute the integral $$\int_0^{e^{-1}}1 dx$$ using the partition $\{e^{-n}, n\geq 1\}$, since the integrand is constant we have

$$\int_0^{e^{-1}}1 dx = \sum_{n=1}^{+\infty}\left(e^{-n} - e^{-(n+1)}\right)$$

i.e. $$e^{-1} = \sum_{n=1}^{+\infty}\left(e^{-n} - e^{-(n+1)}\right) = (1-e^{-1})\sum_{n=1}^{+\infty}e^{-n}$$

so $$\sum_{n=1}^{+\infty}e^{-n} = \dfrac{e^{-1}}{1-e^{-1}} = \frac{1}{e-1}$$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – user642796 Nov 1 '14 at 19:04
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In the following I shall not convert the required sum into a Riemann sum, but use the primitive $x\mapsto -e^{-x}$ of $x\mapsto e^{-x}$ to sum the series without "algebraic tricks". From $$\int_n^{n+1}e^{-x}\>dx=-e^{-x}\biggr|_n^{n+1}=e^{-n}\left(1-{1\over e}\right)\qquad(n\geq1)$$ it follows that $$\int_1^\infty e^{-x}\>dx=\sum_{n=1}^\infty\int_n^{n+1}e^{-x}\>dx=\left(1-{1\over e}\right)\sum_{n=1}^\infty e^{-n}\ .$$ Solving for the sum in question we obtain $$\sum_{n=1}^\infty e^{-n}={e\over e-1}\int_1^\infty e^{-x}\>dx={e\over e-1}\cdot {1\over e}={1\over e-1}\ .$$

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  • $\begingroup$ thank you, but how did you get that integral? How do you get from Riemann Sum to integrals? Please take a look at: math.stackexchange.com/questions/998713/… Where I ask this deeply, perhaps you can help? Thanks! $\endgroup$ – Amad27 Oct 30 '14 at 19:20
  • $\begingroup$ I really like this approach. It's a nontrivial connection between the sum and integration. $\endgroup$ – Joel Nov 3 '14 at 16:21

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