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I've seen this being asked, but I didn't find anything useful for me to solve the problem. I've seen people saying that it can be solved using the equivalence $ (1+ \frac{1}{n})^n = \sum_1^n \frac{1}{k!}$, but I don't get how you can just move things out of the summation and limit to cancel them out (which is what they did to show that the two things are equivalent).

I'm thinking of proving it by proving that the sequence is increasing and bounded above. I can prove that it's bounded above by showing that it's less than $\sum_1^n \frac{1}{k!}$ since, $$\big(1+ \frac{1}{n}\big)^n = \sum_1^n \big(\frac{1}{k!}\big)\big(\frac{n-1}{n}\big)\big(\frac{n-2}{n}\big)\ldots < \sum_1^n \frac{1}{k!}$$and that $\sum_1^n \frac{1}{k!}$ is bounded above (by comparing with the geometric series $(1/2^n$), but I have no idea how to prove it's increasing. What should I do?

haha
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    What about the several posts to show via induction that $(1+\frac 1n)^n$ is increasing but is bounded above? Thus it must converge...... Considering how often this has been asked it's going to be hard to argue this isn't a duplicate. Some justification will have to be made why all the other posts weren't satisfactory for your purposes. – fleablood Oct 23 '22 at 01:05
  • https://math.stackexchange.com/questions/193266/a-question-with-the-sequence-e-n-left1-frac1n-rightn – IrbidMath Oct 23 '22 at 01:52
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    https://math.stackexchange.com/questions/1334832/prove-limit-converges-in-definition-of-e – IrbidMath Oct 23 '22 at 01:53
  • @fleablood I didn't find any post that show via induction when I was searching. If you have links it to those posts it would be great. – haha Oct 23 '22 at 02:30
  • Another relevant Q: https://math.stackexchange.com/questions/389793/ – Jam Oct 23 '22 at 08:06
  • In this answer, it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. This means that $\left(1+\frac1n\right)^n$ is increasing and bounded above by $\left(1+\frac15\right)^{5+1}\lt3$. – robjohn Oct 23 '22 at 10:56

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Let $e_n = (1+\frac{1}{n})^n$. Then we get $$\frac{e_{n+1}}{e_n} = \left(1+\frac{1}{n}\right)\left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n+1} = \frac{n+1}{n}\left(\frac{n^2+2n+1-1}{n^2+2n+1}\right)^{n+1} = \frac{n+1}{n}\left(1-\frac{1}{(n+1)^2}\right)^{n+1} > \frac{n+1}{n}\left(1-(n+1)\frac{1}{(n+1)^2}\right)=\frac{n+1}{n}\frac{n}{n+1}=1,$$ using Bernoulli's inequality. Therefore $(e_n)_{n\in \mathbb{N}}$ is increasing.

Another way of showing boundedness above is by defining $f_n = \left(1+\frac{1}{n}\right)^{n+1}$. We then get $\frac{f_n}{f_{n+1}} > 1$ and thus that $(f_n)_{n\in \mathbb{N}}$ is decreasing by using Bernoulli's inequality in an analogous way.

We then have $e_n < f_n \leq f_1 = 4$ for all $n\in \mathbb{N}$ and thus $(e_n)_{n\in \mathbb{N}}$ is bounded above and converges.

hm1912
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