Would the Series Converge if 2 Were Replaced by e?

[theraptor]

Okay, in my Calc class, we're going through infinite series right now, with respect to convergence and divergence. Well, one problem we had was the one badly formatted below (sorry).

Σ(2^k)(k!)/(k^k)

Now, we determined that this series converged, and that if the two had been a three, it would've diverged. My question is, if the two had been e instead, would the series converge or diverge, and how would one go about determining this? The question was raised in class with the ever so helpful answer of "we can't determine that with the methods that we have gone over thus far." Any help would be appreciated.

 

[Hurkyl]

Hrm, presuambly you used the n-th root test with Stirling's approximation? Convergence problems are usually a lot more difficult at the boundary.

There are improved forms of Stirling's approximation; maybe using one of those would shed light on the problem?

 

[NateTG]

Hmm...
\sum_{k=0}^{\infty} \frac{2^k k!}{k^k}
http://mathworld.wolfram.com/StirlingsApproximation.html
Has an approximation that gives
k! \approx \sqrt{(2n+\frac{1}{3})\pi}k^ke^{-k}
so
\sum_{k=0}^{\infty} \frac{2^k k!}{k^k}\approx \sum_{k=0}^{\infty} \sqrt{(2n+\frac{1}{3})\pi}

which is pretty obviously divergent.

 

[Hurkyl]

Oh, right, you can use the lower bound on n!, and can probably get a formula like Nate's which diverges.

 

[theraptor]

Hmm...
\sum_{k=0}^{\infty} \frac{2^k k!}{k^k}
http://mathworld.wolfram.com/StirlingsApproximation.html
Has an approximation that gives
k! \approx \sqrt{(2n+\frac{1}{3})\pi}k^ke^{-k}
so
\sum_{k=0}^{\infty} \frac{2^k k!}{k^k}\approx \sum_{k=0}^{\infty} \sqrt{(2n+\frac{1}{3})\pi}

which is pretty obviously divergent.

Okay...

First, thank you.

Second, that version is a lot prettier than how I entered it... Have to look into that.

Third, that would've been a decent amount simpler if I had known that alternative to factorials. Yeah, we haven't done that...

*ponders what would happen if he pulled that on a test...*