how to find the limit of the sequence $s_n = (1 + \frac{1}{2n})^n$ using the binomial theroem? Any hints , 
The binomial expansion of this will be $1 + \dfrac 1 2 + \dfrac{n-1}{2n} + \cdots$
but i don't know how can i go from there ?
 A: Binomial Theorem states that: 
(x + y)^r = x^r + rx^(r-1)y + r(r-1)/2!x^(r-2)y^2 + r(r-1)(r-2)/3!x^(r-3)y^3 +.....
substitute x = 1, y = 1/2n, and r = n into the equation:
(1 + 1/2n)^n = 1 + n(1/2n) + n(n-1)/2!(1/2n)^2 + n(n-1)(n-2)/3!(1/2n)^3 +.....
= 1 + 1/2 + 1/2!(1/2)^2(n(n-1)/n^2) + 1/3!(1/2)^3(n(n-1)(n-2)/n^3) + ....
Now taking limit term by term as n--> infinity to get:
1 + 1/2 + 1/2!(1/2)^2 + 1/3!(1/2)^3 +.... = e^(1/2).
A: $$
\left(1+\frac1{2n}\right)^n=\sum_{k=0}^n {n \choose k}\,\frac1{(2n)}=\sum_{k=0}^n \frac{n!}{k!(n-k)!}\,\frac1{n^k}\,\frac1{2^k}=\sum_{k=0}^n \frac{n!}{(n-k)!}\,\frac1{n^k}\,\frac1{k!\,2^k}=\sum_{k=0}^n \left(1-\frac1n\right)\cdots\left(1-\frac{k}n\right)\,\frac1{k!\,2^k}.
$$
Note that 
$$
0\leq1-\left(1-\frac1n\right)\cdots\left(1-\frac{k}n\right)\leq1-\left(1-\frac kn\right)^k
=-\sum_{j=1}^k{k\choose j}(-1)^j\frac{k^j}{n^j}\leq\frac{k^{k+1}k!}n.
$$
Now fix $\varepsilon>0$, and choose $n_0$ such that $k^{k+1}k!/n_0<\varepsilon/4$ and $\sum_{k=n_0+1}^\infty\frac1{k!\,2^k}<\varepsilon/2$. Note also that $1-(1-k/n)^k\leq1$ for all $k\leq n$. Then
$$
\left|\left(1+\frac1{2n}\right)^n-\sum_{k=0}^n\frac{1}{k!\,2^k}\right|\leq\sum_{k=0}^{n_0}
\left(1-\left(1-\frac kn\right)^k\right)\,\frac1{k!\,2^k}+\sum_{k=n_0+1}^n\frac1{k!\,2^k}
\\ \leq\frac\varepsilon4\sum_{k=0}^{n_0}\frac1{2^k}+\frac\varepsilon2\leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon.
$$
So 
$$
\lim_{n\to\infty}\left(1+\frac1{2n}\right)^n=\sum_{k=0}^\infty\frac1{k!\,2^k}=e^{1/2}
$$
