# series

• Apr 15th 2010, 07:13 PM
twofortwo
series
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, 07:35 PM
tonio
Quote:

Originally Posted by twofortwo
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!:)

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

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

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

Tonio
• Apr 16th 2010, 12:09 AM
CaptainBlack
Quote:

Originally Posted by tonio

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

Tonio

????

Cb
• Apr 16th 2010, 03:55 AM
tonio
Quote:

Originally Posted by CaptainBlack
????

Cb

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

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)