# Check whether $\lim\limits_{x \to 1}\sum\limits_{n=0}^{\infty} x^n e^{ - a n} = \frac{1}{1 - e^{-a}}$, for $a>0$

True or false? For any $a > 0$: $$\lim_{x \to 1} \sum_{n=0}^{\infty} x^n e^{ - a n} = \sum_{n=0}^{\infty} e^{- a n} = \frac{1}{1 - e^{-a}}$$

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.

• 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
• Nearly 4 years later, and in the absence of reaction of the asker or the answerer, I might be forgiven to state that the accepted answer does not work, for the reason made explicit in its comments. – Did Jul 20 '17 at 10:58

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: $x^ne^{-an}=e^{n\ln x-an}=e^{n(\ln x -a)}$
• Odd answer, since $x^ne^{-an}$ is dominated by $2e^{-an}$ on NO interval $(1,1+\varepsilon)$. – Did Dec 10 '13 at 18:55
• Hmmm... By the way, even after correction, this is again wrong: choosing $c$ and $x$ like in the post guarantees nothing. – Did Sep 2 '16 at 14:21
Brute force approach: $\color{red}{\text{compute everything!}}$
More precisely: for every $|x|<\mathrm e^a$, $$\sum_{n=0}^{\infty} x^n \mathrm e^{ - a n} =\sum_{n=0}^{\infty} (x \mathrm e^{ - a })^n = \frac1{1-x\mathrm e^{-a}}$$ Now, take the limit $x\to1$.