Determine if series is absolutely convergent

determine if series is absolutely convergent.

sum from 1 to infinity (-6)^n/n!

Here's what I did so far

lim as n approaches infinity (-6)^(n+1)/ (n+1)! x n!/(-6)^n

I did some canceling and got

limit as n approaches infinity absolute value of -6n!/ (n+1)!

When i plug in n for infinity i got a indeterminate form. Answer suppose to be absolutely convergent. what did i do wrong. Help!

You set up that correctly.

So then to make it look better or easier, rewrite it as:

((-6)^(n+1)/(-6)^n) x (n!/(n+1)!)

Then it'll be this:

limit as n approaches infinity -6*(1/(n+1))

We know it's that because if you noticed the (-6), the top has one more than the bottom after you plug in numbers, everything will end up canceling except the -6 on top. Then for the n!/(n+1), if you plug in numbers, you will only be left with 1/(n+1).

And as n approaches infinity, the bottom will get larger and go to 0. And (-6)*(0) is 0.

So absolutely convergent is the answer

Hello,

It is a well known expression that the Taylor series of the exponential function is:
e^x = ∑(n=0→+∞) xⁿ/n!

Since your series is ∑(n=1→+∞) (-6)ⁿ/n!
it's easy to deduce:

∑(n=1→+∞) (-6)ⁿ/n! = [ ∑(n=0→+∞) (-6)ⁿ/n! ] - 1
   = e⁻⁶ - 1

Thus your series is absolutly convergent and its value is (e⁻⁶-1).

Regards,
Dragon.Jade :-)