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?

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!
