when is the sequence: ((-a)^n)/(n!) convergent?

[dts5044]

For what values of the constant a is the sequence a[sub:smhqwpyu]n[/sub:smhqwpyu] = (-a)[sup:smhqwpyu]n[/sup:smhqwpyu] / n! (n>=1) convergent, and what is its limit?

for this one, I tried to use the squeeze theorem. I broke (-a)[sup:smhqwpyu]n[/sup:smhqwpyu] / n! into:

(-a) *(-a) * (-a) .... n times
--------------------------------
n * (n-1) * (n-2) * (n-3)...(n-n +2) * (1)

if a is negative then (-a) is positive, and the previous term is less than its largest factor, which is (-a), and greater than its smallest factor, which is (-a) / n
so I end up with the inequality (-a) / n <= ((-a) *(-a) * (-a) .... n times) / n * (n-1) * (n-2) * (n-3)...(n-n +2) * (1) <= (-a)

taking the limits of either side and using the squeeze theorem I get:
0<= lim n--> infinity ((-a) *(-a) * (-a) .... n times) / n * (n-1) * (n-2) * (n-3)...(n-n +2) * (1) <= (-a)

which proves convergence (since a is fixed), but doesn't tell me what the limit is...

if a is positive then (-a) is negative, and the inequality can be rewritten as:
(-a) / n >= ((-a) *(-a) * (-a) .... n times) / n * (n-1) * (n-2) * (n-3)...(n-n +2) * (1) >= (-a)

...unless n is even, in which case:

(a) / n <= ((-a) *(-a) * (-a) .... n times) / n * (n-1) * (n-2) * (n-3)...(n-n +2) * (1) <= (a)

in any case, the sequence is convergent for any real number a, as long as my work is correct. But I'm stuck with how to get what the limit is. Can anyone help?

 

[Dr. Flim-Flam]

Use the ratio test.

 

[daon]

Remember: for all real x,

\(\displaystyle e^x := \sum _{n \ge 0} \frac{x^n}{n!}\)

hint: This series converges.