The expansion of $a_n=(1+\frac{1}{n})^n$ and the proof of the limit.

Question

I am stuck in the proof of the limit for the sequence $a_n=(1+\frac{1}{n})^n$ where it is shown that the sequence is bounded. Here it is:$$a_n=(1+\frac{1}{n})^n\le1+1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}\le 1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}\le1+\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}}$$ Why is $1+\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}}$ greater than $1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}$?

And why $1+\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}}\le2$? (If $n\ge2)$

Answer 1

Note that$$1+\frac12+\frac1{2^2}+\cdots+\frac1{2^{n-1}}=1+\frac12+\left(\frac12\right)^2+\cdots+\left(\frac12\right)^{n-1}.$$This is the sum of a geometric sequence; its sum is precisely$$\frac{1-\left(\frac12\right)^n}{1-\frac12}=\frac{1-\frac1{2^n}}{1-\frac12}.$$

Answer 2

Another approach to showing that $a_n=(1+\frac{1}{n})^n$ is bounded: $$a_n=(1+\frac{1}{n})^n= \sum_{k=0}^{n}{ {n \choose k} \frac{1}{n^k}}=$$ $$\sum_{k=0}^{n}{ \frac{n!}{k!(n-k)!} \frac{1}{n^k}}=$$ $$\sum_{k=0}^{n}{ \frac{n(n-1)(n-2)...(n-(k-1))}{k!} \frac{1}{n^k}}=$$ $$\sum_{k=0}^{n}{ \frac{n(n-1)(n-2)...(n-(k-1))}{n^k} \frac{1}{k!}} \le \sum_{k=0}^{n}{\frac{1}{k!}}$$

The last step is valid because $\frac{n(n-1)(n-2)...(n-(k-1))}{n^k} \le 1$.
$\sum_{k=0}^{\infty}{\frac{1}{k!}}$ is bounded. This can be shown by noting that $k! > 2^k$ as $n$ grows.

We showed that $a_n=(1+\frac{1}{n})^n \le \sum_{k=0}^{\infty}{\frac{1}{k!}}$ so $a_n$ is bounded.