why is the limit as n goes to infinity of $(1+\frac{1}{n}+\frac{200}{n^2})^n = e$? I know that $$\lim_{n\to\infty}\left(1+\frac{1}n\right)^n = e .$$ 

But why does $$\lim_{n\to\infty}\left(1+\frac{1}n+\frac{a}{n^b}\right)^n = e ? \quad where\quad  b\gt1$$
  better yet, how can I conclude something like:
  $$\lim_{n\to\infty}\left(1+\frac{1}n+\sum_{k=2}^\infty \frac{700^k}{k!n^k}\right)^n = e $$
  Why do all the terms in the sigma not contribute anything to limit? 

This is from a statistics course where we have to evaluate a similar expression but I have studied and done most of the exercises of the chapter on sequences and series of real numbers in Rudin's principles of math. analysis 
 A: If $g(n)$ is  a real valued function to the natural numbers with $n(g(n)-1)\to 0$ then $g(n)^n\to 1$. (Full proof on another question here.)
Then take $$g(n)=\frac{1+\frac1n + \frac a{n^b}}{1+\frac{1}{n}}\; b>1$$
Then we see that $$\left|n(g(n)-1)\right|=\dfrac{\dfrac {|a|}{n^{b-1}}}{1+\dfrac{1}{n}}<\dfrac{|a|}{n^{b-1}}\to 0.$$
So $g(n)^n \rightarrow 1$. That means that $\left(1+\frac{1}{n}+\frac{a}{n^b}\right)^n$ must converge to the same value as $\left(1+\frac{1}{n}\right)^n$, namely to $e$.
This works with all of your examples - take $g(n)$ as your expression divided by $1+\frac{1}{n}$.
More generally, if $h(n)=1+\frac{a}{n}+o\left(\frac{1}n\right)$, (assuming you know "little-$o$" notation[*]) then $$g(n)=\dfrac{h(n)}{1+\frac{a}{n}}$$ has the above property, and thus $h(n)^{n}\to e^{a}$.

[*] The little-$o$ notation is equivalent of $n\left(h(n)-1-\frac{a}{n}\right)\to 0$.
A: Compare each case to $e^{1+\epsilon} = \lim_{n \to \infty}(1 + \frac{1+\epsilon}n)^n$ and show that for any $\epsilon > 0$, there is an $N$ such that the expression in the above limit becomes bigger for any $n > N$. Therefore $e^1 = e$ is the largest number those could possibly converge to. And since they obviously do not converge to something less than $e$, you're done.
A: $$
\left(1+\frac{1}{n}+\frac{a}{n^b}\right)^n=\left(1+\frac{n^{b-1}+a}{n^b}\right)^{\frac{n^b}{n^{b-1}+a} \frac{n^{b-1}+a}{n^b}\dot n}\to \exp\left({\lim\frac{n^{b}+an}{n^b}}\right)=e 
$$
If $b > 1$.
A: HINT:
$$
\left(1+\frac1n+\frac a{n^b}\right)^n=\left[\left(1+\frac{n^{b-1}+a}{n^b}\right)^{n^b}\right]^{1/n^{b-1}}
$$
A: In situations like this loging helps. Write it out as 
$$
e^{n \log (1+ \frac{1}{n} + \frac{200}{n^2})}
$$
You can take the limit 'inside' the exponential function because it is continuous. Now that you've done that, expand $\log( \cdot )$ into Taylor series around 1, which would work in this case becase the limit of the log function inside is $0$ at $n=\infty$: 
$$
e^{1 + \frac{1}{n} + \frac{200}{n}} \to e
$$
A: I normally don't like using the $\log$ function when dealing with limits of this type, because there is a real danger of circularity in arguments. However, if we accept that $\lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e$, and, more generally,  that $\lim_{n \to \infty}( 1 + \frac{x}{n})^{n} = e^{x}$ for any real number $x$, then we have an agreed starting point.
Now the inverse function to the exponential function is the natural logarithm function $\log x$ which has derivative $\frac{1}{x}.$  For any positive real number $x$, the mean value theorem gives $\frac{\log(1+x) - \log(1)}{x} = \frac{1}{1+\theta}$ for some $\theta \in (0,x)$, so that $\log(1+x) < x$.
Since it's more of a challenge, let's look at your second limit. I'll write it as $\lim_{n \to \infty}(1 + \frac{1}{n} + \sigma)^{n}$ for convenience.
Now we have $\log( 1 + \frac{1}{n} + \sigma)^{n} = n \log( 1+ \frac{1}{n} + \sigma) \leq n ( \frac{1}{n} + \sigma ) \leq (1 + n \sigma)$.
It is clear that $\lim_{n \to \infty} n\sigma = 0$ from your formula for $\sigma$.  Hence the log is tending to something at most $1$ as $n \to \infty$. 
However, we know that the limit itself is at least $e$, so the log we are taking is tending to $1$ as $n \to \infty$.
Since $\log$ is a continuous function, and the log of the expression is tending to $1$ as $n \to \infty$, the expression itself must have limit $e$.
A: Fix $a,b \in \mathbf{R}$ with $b>1$. Then for every $\varepsilon>0$
$$
\left(1+\frac{1-\varepsilon}{n}\right)^n \le \left(1+\frac1n+\frac a{n^b}\right)^n \le \left(1+\frac{1+\varepsilon}{n}\right)^n
$$
whenever $n$ is sufficiently large. The claim follows taking the limit of each side.
