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True or false:
For any $a > 0$:

$\displaystyle \lim_{x \to 1} \sum_{n=0}^{\infty} x^n e^{ - a n} = \sum_{n=0}^{\infty} e^{- a n} = \frac{1}{1 - e^{-a}}$

I know when we can exchange limits and integrals, by using the dominated convergence theorem, but I don't know how to deal with exchanging limits and sums.
Any ideas? Thanks.

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You can treat a sum as an integral with respect to the counting measure. –  Brandon Dec 10 '13 at 17:57
    
Thanks Brandon, I used this. The other answer were good too. –  MrReese Dec 10 '13 at 18:09
    
MrReese Were you able to turn the indications in the accepted answer into a full solution? I am curious to see the result... –  Did Dec 10 '13 at 18:56
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2 Answers

up vote 3 down vote accepted

The dominated convergence theorem for series is essentially exactly the same as the one for integrals. In this case, you need $x^ne^{-an}$ to be bounded by something summable, for which you can take something like $e^{-c n}$, where $c=\ln\epsilon-a$ and $x<1+\epsilon$. This is from: $x^ne^{-an}=e^{n\ln x-an}=e^{n(\ln x -a)}$

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Odd answer, since $x^ne^{-an}$ is dominated by $2e^{-an}$ on NO interval $(1,1+\varepsilon)$. –  Did Dec 10 '13 at 18:55
    
@Did: Saw a left sided limit, whoops. Thanks for the correction! –  Alex R. Dec 10 '13 at 20:20
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Brute force approach: compute! $$\sum_{n=0}^{\infty} x^n \mathrm e^{ - a n} =\sum_{n=0}^{\infty} (x \mathrm e^{ - a })^n = \frac1{1-x\mathrm e^{-a}}$$

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