Infinite series convergence question:

[Contingency]

Homework Statement

Does $$\sum _{ n=1 }^{ \infty }{ \frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } } }$$ converge $$\forall |\alpha |<e$$
and if so, how can I prove it?

Homework Equations

$${ e }^{ x }=\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } }$$

The Attempt at a Solution

In the case of e I get the strange and probably wrong $$\sum _{ n=1 }^{ \infty }{ \frac { { (\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } ) }^{ n }{ n }! }{ { n }^{ n } } }$$
which seems to be the boundary between convergence and divergence for the series, but I have no idea how to actually make anything of this.

 

[LCKurtz]

Homework Statement

Does $$\sum _{ n=1 }^{ \infty }{ \frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } } }$$ converge $$\forall |\alpha |<e$$
and if so, how can I prove it?

Homework Equations

$${ e }^{ x }=\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } }$$

The Attempt at a Solution

In the case of e I get the strange and probably wrong $$\sum _{ n=1 }^{ \infty }{ \frac { { (\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } ) }^{ n }{ n }! }{ { n }^{ n } } }$$
which seems to be the boundary between convergence and divergence for the series, but I have no idea how to actually make anything of this.

Have you tried the ratio test for convergence?

 

[Contingency]

Yup.
I tried it on $$\frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } }$$
and ended up with: $$\lim _{ n\rightarrow \infty }{ \frac { { \alpha }^{ n }{ (n+1){ n }^{ n } } }{ { (n+1) }^{ n+1 } } }$$ but I don't see anything relating alpha to e.. I also don't know how to calculate that limit, I'm guessing it's zero but that would mean it's independent of alpha(≠0)..

EDIT: typo correction:
$$\lim _{ n\rightarrow \infty }{ \frac { { \alpha }{ (n+1){ n }^{ n } } }{ { (n+1) }^{ n+1 } } }$$

 

[Contingency]

I'll try to manipulate the expression to get e out of the form of (1+1/n)^n

 

[Contingency]

ah, I did the above and my result is that for any alpha>e the ratio test yields r>1 and for any alpha<e the ratio test yields r<1. I'll think about r=1

 

[LCKurtz]

Yup.
I tried it on $$\frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } }$$
and ended up with: $$\lim _{ n\rightarrow \infty }{ \frac { { \alpha }^{ n }{ (n+1){ n }^{ n } } }{ { (n+1) }^{ n+1 } } }$$ but I don't see anything relating alpha to e.. I also don't know how to calculate that limit, I'm guessing it's zero but that would mean it's independent of alpha(≠0)..

You need absolute values around that, and I don't think you get $\alpha^n$ after you simplify the ratio. Check that.

 

[Ray Vickson]

Homework Statement

Does $$\sum _{ n=1 }^{ \infty }{ \frac { { \alpha }^{ n }{ n }! }{ { n }^{ n } } }$$ converge $$\forall |\alpha |<e$$
and if so, how can I prove it?

Homework Equations

$${ e }^{ x }=\sum _{ n=0 }^{ \infty }{ \frac { { x }^{ n } }{ n! } }$$

The Attempt at a Solution

In the case of e I get the strange and probably wrong $$\sum _{ n=1 }^{ \infty }{ \frac { { (\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n! } } ) }^{ n }{ n }! }{ { n }^{ n } } }$$
which seems to be the boundary between convergence and divergence for the series, but I have no idea how to actually make anything of this.

Use Stirling's formula for the asymptotic form of n!. You can even improve it to a rigorous inequality that holds for all n > 1 (even small values of n):
$$\sqrt{2\pi n} n^n e^{-n} < n! < \sqrt{2\pi n} n^n e^{\left(-n + \frac{1}{12n}\right)}.$$
Note: the left-hand quantity above is Stirling's asymptotic approximation to n!; the right-hand quantity is an improvement, and it is surprising just how good it is: for n = 2 we get 2! = 2, but RHS = 2.000652048, 5! = 120, RHS = 120.0026371, etc. Various "simple" proofs of these inequalities are available; my favourite one is in Feller, "Introduction to Probability Theory", Vol. 1, Wiley 1968.

RGV

 

[Contingency]

@MLCKurtz
Yes, I obtained alpha without an exponent, made a typo. Didn't forget the absolute value. Thanks
@Ray Vickson
Thanks for that reading material! I think i'll stick to my simpler idea, considering I don't have justifications to apply Stirling's formula (haven't gotten around to it yet).