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We have $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!} $$ and $$\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^n \frac1{n^k}{n\choose k}x^k $$ therefore $$\left\lvert e^x-\left(1+\frac{x}{n}\right)^n\right\rvert\le\sum_{k=n+1}^\infty \frac{\lvert x\rvert^k}{k!}+\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert. $$

Question: As $n\to\infty$, the first sum trivially goes to $0$, but how to properly bound the second one?

I can show convergence in different ways but I'm interested in this specific approach.

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    $\begingroup$ Interesting question! To avoid getting other methods, I would specifically write down: **Question **: how can one show that the sum (the second sum in your question) goes to zero? (does it even go to zero?). Because there are many things one can do before the triangle inequality step to prove the claim, but that boils down doing the classic proof. That is what I think at least. $\endgroup$ – Shashi Dec 12 '17 at 10:23
  • $\begingroup$ This is one of the standard ways of proving the equality of the exponential limit and the exponential series and if you search enough you will find this approach in detail on this website. $\endgroup$ – Paramanand Singh Dec 12 '17 at 12:41
  • $\begingroup$ @Shashi I've followed your advice, thanks $\endgroup$ – Richard Dec 12 '17 at 18:10
  • $\begingroup$ See this answer math.stackexchange.com/a/1898375/72031 $\endgroup$ – Paramanand Singh Dec 16 '17 at 7:04
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\begin{align}\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert &\le\sum_{k=1}^{\infty}\frac{\lvert x\rvert^k}{k!} \underbrace{\left\lvert \big(\tfrac{n\cdot(n-1)\cdot\ldots\cdot(n-(k-1))}{n^k }- 1\big) \right\rvert}_{:=a_k(n)} \end{align}

Now let $\epsilon>0$ be arbitrary.

On the one hand $0\le a_k(n)\le 1$ for all $k$, hence the above series is bounded by $\exp(|x|)-1$. In particular there exists some $N$ such that $\sum_{k=N}^\infty a_k(n)\frac{|x|^k}{k!} \le \sum_{k=N}^\infty \frac{|x|^k}{k!} < \epsilon$.

On the other hand, since $a_k(n) \to 0$ for all $k$ we can then also find some $n_\epsilon$ such that $a_k(n_\epsilon) < \epsilon$ for all $k<N$. Putting the two together we conclude

\begin{align} \sum_{k=1}^{\infty}a_k(n_\epsilon)\frac{|x|^k}{k!} \le \sum_{k=1}^{N-1}\epsilon\frac{|x|^k}{k!} + \sum_{k=N}^{\infty}\frac{|x|^k}{k!} \le \epsilon(\exp(|x|)-1) + \epsilon = \exp(|x|)\epsilon \end{align}

Which demonstrates point-wise convergence.

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  • $\begingroup$ Thanks! +1, I understand and appreciate your justification. I was wondering, since the infinite sum at the beginning converges for all $n$ and doesn't depend on $n$, couldn't we also directly take the limit as $n\to\infty $ and get $0$ as $a_k(n)\to 0$ for all $k $? $\endgroup$ – Richard Dec 12 '17 at 13:52
  • $\begingroup$ @Richard I did that in an earlier answer that I deleted; user Shashi rightly pointed out that the swapping of the limit with the sum needs some justification. This could be done by a uniform convergence argument, but I then rather chose to give the elementary approach that doesn't need any more machinery. $\endgroup$ – Hyperplane Dec 12 '17 at 15:36
  • $\begingroup$ Didn't you do that when the sum still depended on $n$? I mean, the infinite sum does not, so taking the limit there should be safe, right? $\endgroup$ – Richard Dec 12 '17 at 17:24
  • $\begingroup$ To me it doesn't look different from $\sum_{k=0}^\infty \frac{x^k }{(n+k)!} \to 0$ as $n\to\infty $ $\endgroup$ – Richard Dec 12 '17 at 17:32
  • $\begingroup$ Sorry to bother you, could you please consider my doubt? $\endgroup$ – Richard Dec 13 '17 at 18:13
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All the coefficients in the 2nd sum are non-positive. Indeed,

\begin{multline*}\frac{1}{n^k}\binom{n}{k}-\frac{1}{k!}=\frac{1}{k!}\left(\frac{n!}{n^k(n-k)!}-1\right)=\frac{1}{k!}\left(\frac{(n-k+1)(n-k+2)\cdot\ldots\cdot n}{n^k}-1\right)\\\le \frac{1}{k!}\left(\frac{n^k}{n^k}-1\right) = 0.\end{multline*}

Performing your computation without absolute value and observing that for $x>0$ we have the difference on the left is positive, we have done for $x>0$. It seems to me that for $x<0$ the similar argument will work (with the reverted inequalities). Please check it. For $x=0$ the checking is direct.

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