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Problem: Given a convergent sequence $x_{n} \to x_{0}$ as $n \to \infty$, and that e is defined as $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$, prove that $\lim_{n \to \infty}e^{x_{n}} = e^{x_{0}}$.

Now I know I could simply use limit rules, and say that $\lim_{n \to \infty}e^{x_{n}} = e^{\lim_{n \to \infty}x_{n}} = e^{x_{0}}$. However I would like (and the question suggests) to use that infinite series definition of $e^{x}$ to arrive at the answer.

So far I have $\lim_{n \to \infty}e^{x_{n}} = \lim_{n \to \infty}\sum_{k=0}^{\infty}\frac{x_{n}^k}{k!} =\sum_{k=0}^{\infty}\frac{x_{0}^k}{k!}$, but again without just distributing the limit inside the summation, I am stuck as to how to proceed from there and find the limit of that series. Any help would be appreciated! (It could be there is no way to do it without simply distributing the limit, I'm not sure)

user1234544
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2 Answers2

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I will offer a solution using the series. Since $x_n\to x_0$, we may assume there exists $M>0$ such that $|x_n|\leq M$ for all $n$. Therefore, $|\sum_{k=0}^\infty \frac{x_n^k}{k!}|\leq \sum_{k=0}^\infty \frac{M^k}{k!}$. Hence, by dominated convergence theorem (with point mass measure on the naturals), you can bring the limit inside the sum of $\sum_{k=0}^\infty \frac{x_n^k}{k!}$ so $\lim_{n\to \infty}e^{x_n}=e^{\lim_{n\to\infty} x_n}=e^{x_0}$.

ProfOak
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  • zugzug.Please clarify.Hour argument shows that $\sum x^k_n/k!$ is bounded, ok. Uniformly convergent? Do I get something wrong? – Peter Szilas Mar 13 '20 at 20:03
  • It's really just dominated convergence theorem. The sum can be viewed as an integral with respect to dirac point mass measure. – ProfOak Mar 13 '20 at 20:31
  • It's the DCT, yes but nothing to do with uniform convergence, since the sums $S_N=|\sum_{k=0}^N \frac{x_n^k}{k!}|$ are numerical sequences. – Matematleta Mar 13 '20 at 23:19
  • @Matematleta I edited it for clarity. – ProfOak Mar 13 '20 at 23:38
  • I was happy to upvote your correct proof. Still, it's overkill. I think the one I offered is easier and more elementary. – Matematleta Mar 14 '20 at 01:11
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Suppose we can prove that $x_k\to 0\Rightarrow e^{x_k}\to 1.$ Then, if $x_k\to x_0,\ y_k:=x_k-x_0\to 0$ and then $e^{y_k}=e^{x_k-x_0}\to 1$ and this implies that $e^{x_k}\to e^{x_0}$.

So, it suffices to prove the result for $x_0= 0.$ But this is easy: choose $K$ large enough so that $k>K\Rightarrow |x_k|<\epsilon<1.$ Then, for such $k$

$ \left|\sum_{k=0}^\infty \frac{x_n^k}{k!} - 1\right|= \left|\sum_{k=1}^\infty \frac{x_n^k}{k!}\right|\le \sum_{k=1}^\infty \frac{\epsilon^k}{k!}\le\sum_{k=1}^\infty \epsilon^k=\frac{1}{1-\epsilon}-1$

The result follows (since $\epsilon$ is arbitrary).

Matematleta
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  • +1 You have implicitly used the result that power series for $e^x $ directly implies the identity $e^{x+y} =e^xe^y$. This is also my preferred approach for this question. If one is aware of power series stuff then one can directly say that they allow limit passing, term by term differentiation and integration in the interior of region of convergence. – Paramanand Singh Mar 14 '20 at 02:41
  • Yes, indeed. I assumed that if the OP was using the series definition for the exponential function, then he/she had proved the basic properties. In any case, it's an easy application of the theorem on Cauchy products. – Matematleta Mar 14 '20 at 02:55