Hello,

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

Thank you

Printable View

- Mar 29th 2010, 11:31 AMmohammadfawazSeries Divergence
Hello,

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

Thank you - Mar 29th 2010, 11:36 AMLaurent
- Mar 29th 2010, 11:47 AMmohammadfawaz
What I actually did is to use the binomial theorem.

We get an inequality for $\displaystyle (1+\frac{1}{k})^k$:

$\displaystyle (1+\frac{1}{k})^k<1+1+\frac{1}{2!}+...+\frac{1}{k! }$.

Expanding e, and taking the difference, I got:

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

But I'm stuck there. I don't know if I'm on the right track.

Thank you - Mar 29th 2010, 12:15 PMLaurent
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 $\displaystyle \log(1+x)\geq x-\frac{x^2}{2}$ for $\displaystyle x\geq 0$ in the expansion I wrote. Or the asymptotic expansion $\displaystyle \log(1+x)=x-\frac{x^2}{2}+o(x^2)$ as $\displaystyle x\to 0$.