1. ## 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!

2. 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? $\displaystyle \sqrt[n]{\frac{n2^n}{n!}}=\frac{\sqrt[n]{n}2}{\sqrt[n]{n!}}\xrightarrow [n\to\infty]{}0<1\Longrightarrow$ 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

3. Originally Posted by tonio

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
????

Cb

4. Originally Posted by CaptainBlack
????

Cb

Perhaps I missed something (nothing too surprising, uh? ), 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...