# Series Divergence

• Mar 29th 2010, 12:31 PM
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
• Mar 29th 2010, 12:36 PM
Laurent
Quote:

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...
• Mar 29th 2010, 12:47 PM
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
• Mar 29th 2010, 01:15 PM
Laurent
Quote:

$e-(1+\frac{1}{k})^k>\frac{1}{(k+1)!}+\frac{1}{(k+2)! }+...$
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 $x\to 0$.