Find sum of series 1

Hong Quy · Yang-Mills Theory Offline Joined: 20 Feb 2005 Posts: 543

Caculate sum of n term of the series: $\sum_{n=1}^{\infty} k e^{k-1}$

**Posted:** Sat Oct 31, 2009 3:11 am

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Stank

Well first of all, I'm assuming you meant $\sum_{k=1}^{\infty}k e^{k-1}$ , not $\sum_{n=1}^{\infty}k e^{k-1}$ . Just looking at this, my math instincts tell me that it diverges (and looking at the first few terms this seems pretty obvious). We can apply the integral test to determine this for sure (see http://mathworld.wolfram.com/IntegralTest.html). $\int k e^{k-1}\,dk=e^{k-1}(k-1)$ . We know that $\displaystyle\lim_{k\to\infty}e^{k-1}(k-1)=\infty$ . Since $\int^\infty_1 k e^{k-1}\,dx$ diverges, it immediately follows that $\sum_{k=1}^{\infty}k e^{k-1}$ also diverges. Hope this was helpful.

**Posted:** Sat Oct 31, 2009 12:36 pm

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JBL

Probably Hong Quy meant to give the non-ridiculous question of computing $\sum_{k = 1}^n k e^{k - 1}$ . (Also, the integral test is ridiculous overkill: the terms of the series are diverging to $+\infty$ Rolling Eyes

)

**Posted:** Sat Oct 31, 2009 1:01 pm

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Joel
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Stank

Yeah, I guess. Smile

Just wanted to prove it.

**Posted:** Sat Oct 31, 2009 1:10 pm

JBL

And my point is that you don't need the integral test to prove that this sum diverges: it fails the $n$ -th term test (because its individual terms do not go to zero, and in fact go to infinity).

Probably the easiest way to solve this problem is to first integrate with respect to $e$ Wink

But there are plenty of elementary approaches as well, and it doesn't really have anything to do with calculus.

**Posted:** Sat Oct 31, 2009 1:13 pm

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Joel
Hi Deeps! <3