The limit of $\lim_{n \to \infty}\left(1 + \frac{1}{\sqrt{n}}\right)^n$ I am approaching the question from an inequality perspective. In other words, I just want to see if the equation has an upper bound or lower bound.
After expanding the equation using binomial expansion, I get the term that $$\left(1 + \frac{1}{\sqrt{n}}\right)^n \leq 1 + n^{1/2} + \frac{n}{2} + \frac{n^{3/2}}{2^{2}}$$
For $n^{1/2} + \frac{n}{2} + ...$ $$a = \sqrt{n}, r = \frac{\sqrt{n}}{2}$$I then use the sum to infinity formula $S_{\infty} = \frac{a}{1 - r}$to get $\frac{4{\sqrt{n}} + 2n}{4 - n}$, add back in the 1 and simplify to get $\frac{\frac{4}{n} + \frac{4}{\sqrt{n}} + 1}{\frac{4}{n} - 1}$.
Finally, when I apply the limit of n to infinity, I get back -1. But, this does not seem right to me. Looking back at the equation, if n is positive, the sum to infinity should be a positive number instead. 
My guess is that the ratio that I used when calculating sum to infinity is wrong. The ratio should be less than 1, but my ratio is more than 1 if n tends to infinity. 
How do I go about solving this?
 A: Note that $\lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^{\sqrt n}=e>2$ so that for $n$ large enough we have $\left(1+\frac1{\sqrt n}\right)^{\sqrt n}>2$ and consequently: $$\left(1+\frac1{\sqrt n}\right)^{n}>2^{\sqrt n}$$
This shows that: $$\lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^{n}=+\infty$$
A: Let $n=k^2$ then
$$\lim_{n \to \infty}(1 + \frac{1}{\sqrt{n}})^n=\lim_{k \to \infty}(1 + \frac{1}{k})^{k^2}=\lim_{k \to \infty}\left((1 + \frac{1}{k})^{k}\right)^k\to e^\infty=\infty$$
A: By Bernoulli inequality $(1+x)^r\ge 1+rx$ we have
$$\left(1 + \frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\ge1 + \sqrt{n}\frac{1}{\sqrt{n}}=2$$
and therefore
$$\left(1 + \frac{1}{\sqrt{n}}\right)^{{n}}=\left[\left(1 + \frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right]^{\sqrt{n}}\ge2^{\sqrt{n}}\to \infty$$

As an alternative by Taylor's expansion we have
$$\left(1 + \frac{1}{\sqrt{n}}\right)^{n}=e^{n\log \left(1 + \frac{1}{\sqrt{n}}\right)}=e^{n\left(\frac{1}{\sqrt{n}}+O(1/n)\right)}=e^{\left(\sqrt n+O(1)\right)}\sim e^\sqrt n\to \infty$$
A: You can use the binomial expansion of the equation:
$$\left(1 + \frac{1}{\sqrt{n}}\right)^n = \sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$
Since all terms are positives, you can fix a lower bound:
$$\sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} \ge \sum_{k=n-1}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$
Then note that:
$$\sum_{k=n-1}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} =$$$$ \begin{pmatrix} n\\n-1  \end{pmatrix}  1^{n-1}\times\left( \frac{1}{\sqrt{n}}\right)^1 + \begin{pmatrix} n\\n  \end{pmatrix} 1^{n} \times\left(\frac{1}{\sqrt{n}}\right)^0 \\
$$$$=n\times\frac{1}{\sqrt{n}} + 1$$
So, you have 
$$\left(1 + \frac{1}{\sqrt{n}}\right)^n \ge 1+\sqrt{n}$$
From there, you can deduce your limit!
A: So, basically you are saying:
$$(1 + \frac{1}{\sqrt{n}})^n\le \frac{\frac{4}{n} + \frac{4}{\sqrt{n}} + 1}{\frac{4}{n} - 1},$$
however, it is not true by WA for $n\ge 5$. It implies that your original estimate with the geometric progression is not valid.
