Let $a_n =\left (1+\frac{1}{n}\right)^n$ and $b_n =1+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}$, given $n$ find $k$ so that $a_k \gt b_n$ Let $a_n = \left(1+\frac{1}{n}\right)^n$ and $b_n = 1 + \frac{1}{1!} + \frac{1}{2!} +\cdots + \frac{1}{n!}$.
I need to find, for every $n$, a $k$ such that  $a_k \gt b_n$.
I proved, that $b_n \ge a_n$ for every $n$, and both sequences are increasing, and $\lim_{n\to \infty} a_n = e$, but I can't use,that $\lim_{n\to \infty} b_n = e$ and i can't use Taylor series.
Could anyone give me a hint? I have no idea about approaching this...
 A: Expand using the binomial theorem:
$$a_k = \sum_{i=0}^{k} \binom{k}{i}\frac{1}{k^i}$$
Recall that $\binom{k}{i}=\frac{k(k-1)\cdots(k-(i-1))}{i!}$, so:
$$\binom{k}{i}\frac{1}{k^i}=\frac{(1-0/k)(1-1/k)\cdots(1-(i-1)/k)}{i!}$$
Given $n$, define the polynomial:
$$p_n(x)=\sum_{i=0}^{n+1}\frac{(1-x)(1-2x)(1-3x)\cdots(1-(i-1)x)}{i!}$$
Then we have that $p_n(1/k)< a_k$ when $k>n+1$, and $\lim_{x\to 0} p_n(x) =p_n(0)=b_{n+1}>b_n.$
So $p_n$ is continuous, we can find $\delta>0$ such that when $|x|<\delta$, $|p_n(x)-b_{n+1}|<\frac{1}{(n+1)!}=b_{n+1}-b_n.$ 
Choose some $k>\frac{1}{\delta}.$
Then $\left|\frac1k\right|<\delta$ and thus $b_n=b_{n+1}-\frac{1}{(n+1)!}<p_n\left(\frac 1k\right)$. If we pick $k>n$, we also get:
$$b_n < p_n(1/k) < a_k$$
It's not coming to me immediately how we can find a specific $k$, without some pain. 
If $p_n(x)=\sum_{k=0}^{n} q_ix^{i}$ let $M=\sum_{i=1}^{n+1}|q_i|$. Then $$p_n(x)-p_n(0)=x\left(\sum_{i=1}^{n+1} q_ix^{i-1}\right)$$ If $|x|<1$ then $|p_n(x)-p_n(0)|<M|x|$.  So if $\delta=\min(\frac{1}{M(n+1)!},1)$ you can choose $k>\max(1,M(n+1)!)$. Just need an upper bound on $M$ now.
But $p_{n}(-1)=\sum_{i=0}^{n+1} |q_i|$, and $q_0=b_{n+1}$, so we see that $$p_{n}(-1)=1+\sum_{i=1}^{n+1} \frac{i!}{i!}=n+2$$ Also, $q_0=p_n(0)=b_{n+1}$. So:
$$M=p_{n}(-1)-p_n(0)=n+2-b_{n+1}<n$$
So we can choose $$k>n(n+1)!$$
A: $b_n$ is the Taylor series of $f(x) = \exp(x)$ (expanded about x=0) evaluated at $x = 1$ with n terms. [This is just background information, not stricty required.]
We have that  $b_n$ will approach e from below. A simple estimate is 
$$
b_n + \frac{1}{(n+1)!} < e
$$
hence $b_n < e -  \frac{1}{(n+1)!}$. So it is enough to find a $k$ such that $e -  \frac{1}{(n+1)!} < a_k = (1 + 1/k)^k$.
This is equivalent to $\log(e -  \frac{1}{(n+1)!}) < k \log (1 + 1/k)$. 
Since $\log(1+x) > x - x^2/2$ (proof see e.g. here), an even harder requirement is 
$$
\log( e -  \frac{1}{(n+1)!}) < 1 - \frac{1}{2\; k}
$$
This finally gives 
$$
k > 
\frac{-1}{2 \log( 1 -  \frac{1}{e (n+1)!}))}
$$
This is not a nice (or the best) formula, but it shows that a $k$ can always be found. 
A: Using the Taylor series, we get
$$
\sum_{k=0}^n\frac1{k!}\lt e-\frac1{(n+1)!}
$$
Using the Taylor series of the log, we get
$$
\left(1+\frac1n\right)^n\gt e-\frac e{2n}
$$
Therefore,
$$
\left(1+\frac1k\right)^k\gt e-\frac e{2k}\gt e-\frac1{(n+1)!}\gt\sum_{j=0}^n\frac1{j!}
$$
That is, we want
$$
k\gt\frac e2(n+1)!
$$
