# Thread: Series Divergence

1. ## Series Divergence

Hello,

How do I show that the series $\sum_{k=1}^\infty [e-(1+\frac{1}{k})^k]$ diverges. The question asks to us to use Maclaurin series to compare the series with some other series.

Thank you

2. Originally Posted by mohammadfawaz
The question asks to us to use Maclaurin series to compare the series with some other series.
So how did it work for you? I guess you wrote $\left(1+\frac{1}{k}\right)^k = \exp\left(k\log\left(1+\frac{1}{k}\right)\right)$ and expanded the exponent first, up to some order, etc...

3. What I actually did is to use the binomial theorem.
We get an inequality for $(1+\frac{1}{k})^k$:
$(1+\frac{1}{k})^k<1+1+\frac{1}{2!}+...+\frac{1}{k! }$.
Expanding e, and taking the difference, I got:
$e-(1+\frac{1}{k})^k>\frac{1}{(k+1)!}+\frac{1}{(k+2)! }+...$
But I'm stuck there. I don't know if I'm on the right track.

Thank you

4. Originally Posted by mohammadfawaz
$e-(1+\frac{1}{k})^k>\frac{1}{(k+1)!}+\frac{1}{(k+2)! }+...$
Good idea, but the right hand side will give you a convergent series, hence you get nothing. (Greater than a convergent series proves nothing)

What you could do is use $\log(1+x)\geq x-\frac{x^2}{2}$ for $x\geq 0$ in the expansion I wrote. Or the asymptotic expansion $\log(1+x)=x-\frac{x^2}{2}+o(x^2)$ as $x\to 0$.