does this series converge or diverge?

n.2^n/n!

i did the limit test and simplified it to

2^n+1/2^n

after this point, im stuck

thanx for the help!:)

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- Apr 15th 2010, 06:13 PMtwofortwoseries
does this series converge or diverge?

n.2^n/n!

i did the limit test and simplified it to

2^n+1/2^n

after this point, im stuck

thanx for the help!:) - Apr 15th 2010, 06:35 PMtonio

What about the n-th root test? $\displaystyle \sqrt[n]{\frac{n2^n}{n!}}=\frac{\sqrt[n]{n}2}{\sqrt[n]{n!}}\xrightarrow [n\to\infty]{}0<1\Longrightarrow $ the series converges.

Or the quotient test: $\displaystyle \frac{(n+1)2^{n+1}}{(n+1)!}\,\frac{n!}{n2^n}=$ $\displaystyle \frac{2}{n}\xrightarrow [n\to\infty]{}0\Longrightarrow $ the series converges.

Or the comparison test: $\displaystyle \frac{n2^n}{n!}\geq \frac{1}{(n-1)!}$ and since $\displaystyle \sum^\infty_{n=1}\frac{1}{(n-1)!}$ converges so does our series.

Tonio - Apr 15th 2010, 11:09 PMCaptainBlack
- Apr 16th 2010, 02:55 AMtonio

Perhaps I missed something (nothing too surprising, uh?(Giggle) ), but what's not clear here? In fact, $\displaystyle \sum^\infty_{n=1}\frac{1}{(n-1)!}=e$ ...

Oh, I see....hehe! Of course, I used the comparisong test in the wrong direction...*sigh*

Well, the OP will have to disregrad that one

Tonio

Ps I'm gonna add a disclaimer to my posts (this is serious, when I'll find out how to do it): "The poster of the above reply has not given too seriously a thought to the OP, so any mistake resulting from the answer is your own unique and complete responsibility".

That'll make it...(Cool)