# how to solve this problem?

• May 20th 2009, 08:21 AM
chris.w
how to solve this problem?
Attachment 11506
• May 20th 2009, 08:54 AM
Kaitosan
1. I believe it's zero because e-e=0....

There's a segment of the series that equals e.

I don't know about 2. though.
• May 20th 2009, 10:02 AM
TwistedOne151
I'm prette sure the first sum diverges. One can check that
$e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for integer $k≥3$.
Now, the harmonic series
$\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, and so
$\sum_{n=3}^{\infty}\frac{1}{n}$ diverges as well.
but since $e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for $k≥3$
then
$\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]\gt\sum_{n=3}^{\infty}\frac{1}{k}$
and so by the comparison test, both sums diverge, and since
$\sum_{n=1}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]=(e-2)+(e-\frac{9}{4})+\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]$,
your series in problem 1 diverges.

--Kevin C.