Showing two sequences converge and have the same limit 
Show that the sequences $$x_n=\sum_{k=0}^n\frac1{k!}\quad\text{and}\quad y_n=\left(1+\frac1n\right)^n$$ are convergent and have the same limit.

I do not know how to show that either of these are bounded sequences, but they are both pretty clearly monotone.
 A: For $k \ge 2,k! > 2^{k-1}$ so
$$\frac{1}{k!} \le \frac{1}{2^{k-1}}$$
i.e. terms of the first sum are all (after a while) bounded by 1/2 + 1/4 + 1/8 +  ... = 1.
A: Although I have provided a link to an answer (of a more general question) in the comments, it would be better to give you the main ideas related to your specific question.
Note that the sequence $x_{n}$ is actually the $n$-th partial sum of a series which is convergent (easily proved via ratio test). Another approach to proving that $x_{n}$ converges is to show that it increases as $n$ increases and is bounded. Clearly as $n$ increase the number of terms in sum defining $x_{n}$ increases and hence the sequence $x_{n}$ increases. Again we can see that if $n > 2$ then $$\begin{aligned}x_{n} &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}\\
&\leq 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{n - 1}}\\
&= 1 + \dfrac{1 - \dfrac{1}{2^{n}}}{1 - \dfrac{1}{2}}\text{ (sum of a finite GP)}\\
&\leq 1 + \dfrac{1}{1 - \dfrac{1}{2}} = 3\end{aligned}$$ Hence $x_{n}$ is bounded and therefore $x_{n}$ tends to a limit.
Next we can focus on $y_{n}$. Clearly if $n > 3$ then we can see that $$\begin{aligned}y_{n} &= \left(1 + \frac{1}{n}\right)^{n}\\
&= 1 + n\cdot\frac{1}{n} + \frac{n(n - 1)}{2!}\left(\frac{1}{n}\right)^{2} + \cdots + \text{ (upto }n\text{ terms)}\\
&= 1 + 1 + \dfrac{1 - \dfrac{1}{n}}{2!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!} + \cdots\end{aligned}$$
Now it is easy to see that as $n$ increases number of terms in the above expression increases as well as each term of the above sum also increases. It follows that $y_{n}$ increases as $n$ increases. Again we can note that $$\begin{aligned}y_{n} &= 1 + 1 + \dfrac{1 - \dfrac{1}{n}}{2!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!} + \cdots\\
&\leq 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}\\
&= x_{n} \leq 3 \text{ (as proved before)}\end{aligned}$$ therefore $y_{n}$ is also bounded and $\lim_{n \to \infty}y_{n}$ also exists.
Also we have seen above that $y_{n} \leq x_{n}$ so that $\lim_{n \to \infty}y_{n} \leq \lim_{n \to \infty}x_{n}$. To establish equality we need to bound $x_{n}$ by another sequence. That sequence is given by $z_{n} = (1 - (1/n))^{-n}$. Clearly using general binomial theorem we have $$\begin{aligned}z_{n} &= 1 + n\cdot\frac{1}{n} + \frac{n(n + 1)}{2!}\left(\frac{1}{n}\right)^{2} + \cdots\\
&= 1 + 1 + \dfrac{1 + \dfrac{1}{n}}{2!} + \dfrac{\left(1 + \dfrac{1}{n}\right)\left(1 + \dfrac{2}{n}\right)}{3!} + \cdots\\
&\geq 1 + 1 + \frac{1}{2!} + \cdots + \frac{1}{n!} = x_{n}\end{aligned}$$ and thus we finally arrive at $$y_{n} \leq x_{n} \leq z_{n}$$ The next thing is to show that both $y_{n}$ and $z_{n}$ tend to same limit as $n \to \infty$ and then by Squeeze theorem $x_{n}$ would also tend to same limit. To proceed further we can note that $$\begin{aligned}\frac{y_{n}}{z_{n}} &= \left(1 - \frac{1}{n^{2}}\right)^{n}\\
&= 1 - n\cdot\frac{1}{n^{2}} + \frac{n(n - 1)}{2!}\left(\frac{1}{n^{2}}\right)^{2} - \cdots + \text{ (upto }n\text{ terms)}\\
&= 1 - \frac{1}{n} + \dfrac{1 - \dfrac{1}{n}}{2!}\left(\frac{1}{n}\right)^{2} - \cdots\\
&= 1 - \phi(n)\end{aligned}$$ where the function $\phi(n)$ is a finite sum defined by $$\phi(n) = \frac{1}{n} - \dfrac{1 - \dfrac{1}{n}}{2!}\left(\frac{1}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{1}{n}\right)^{3}- \cdots + $$ Now we can estimate $\phi(n)$ as $$\begin{aligned}0\leq|\phi(n)| &\leq \frac{1}{n} + \dfrac{1 - \dfrac{1}{n}}{2!}\left(\frac{1}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{1}{n}\right)^{3} + \cdots\\
&\leq \frac{1}{n} + \frac{1}{2!}\cdot\frac{1}{n^{2}} + \cdots\\
&\leq \frac{1}{n} + \frac{1}{2n^{2}} + \frac{1}{2^{2}n^{3}} + \cdots\\
&= \frac{1}{n}\left(1 + \frac{1}{2n} + \frac{1}{(2n)^{2}} + \cdots\right)\\
&= \frac{1}{n}\cdot\dfrac{1 - \dfrac{1}{(2n)^{n}}}{1 - \dfrac{1}{2n}}\\
&\leq \frac{1}{n}\cdot\dfrac{1}{1 - \dfrac{1}{2n}} \to 0\text{ (as }n \to \infty)\end{aligned}$$
It follows that $\lim_{n \to \infty}\phi(n) = 0$ and hence $\lim_{n \to \infty}y_{n}/z_{n} = 1$ and hence $\lim_{n \to \infty}y_{n} = \lim_{n \to \infty}z_{n}$.
