i have been thinking for a long time , please help me

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- May 20th 2009, 08:21 AMchris.whow to solve this problem?
i have been thinking for a long time , please help me

Attachment 11506 - May 20th 2009, 08:54 AMKaitosan
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 AMTwistedOne151
I'm prette sure the first sum diverges. One can check that

$\displaystyle e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for integer $\displaystyle k≥3$.

Now, the harmonic series

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ diverges, and so

$\displaystyle \sum_{n=3}^{\infty}\frac{1}{n}$ diverges as well.

but since $\displaystyle e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k}$ for $\displaystyle k≥3$

then

$\displaystyle \sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]\gt\sum_{n=3}^{\infty}\frac{1}{k}$

and so by the comparison test, both sums diverge, and since

$\displaystyle \sum_{n=1}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]=(e-2)+(e-\frac{9}{4})+\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]$,

your series in problem 1 diverges.

--Kevin C.