what is the limit of the series as n approaches infinity of a(n) = (-3)^n / n! ? i know if it was +3 the limit approaches 0 using the squeeze theorem

Question

terenzreignz · Answer

You can squeeze this too, you know ;) 
$$\huge \frac{-3^n}{n!}\le\frac{(-3)^n}{n!}\le\frac{3^n}{n!}$$

anonymous · Answer

so whats the final answer then?

terenzreignz · Answer

You tell me~ It's always in the middle of two sequences... BOTH of which go to zero.  What do you think? :)

anonymous · Answer

how did u evaluate that -3^n/n! is equal to 0

terenzreignz · Answer

Because
$$\huge \frac{3^n}{n!}\rightarrow 0$$Then if you take its negative, it would go to the negative of the limit, but then again, -0 = 0

anonymous · Answer

oh ok so i did not know u could take the negative of the limit ;)

anonymous · Answer

what if u made the left hand inequality just 0, would it be valid

terenzreignz · Answer

No, because sometimes, your sequence is less than zero.
You need  a sequence that is less than that at every turn.

anonymous · Answer

oh ok now i understand, thanks a lot man :)

terenzreignz · Answer

No problem :)

anonymous · Answer

just 1 more query before i close this question, can u please post your method of evaluating the limit of 3^n/n! is equal to 0. I want to see if it is simpler than the one taught to me

terenzreignz · Answer

Sorry, i was occupied @babycry 
hang on...

terenzreignz · Answer

Do you mean an epsilon proof?

terenzreignz · Answer

Coz I don't have one :)

anonymous · Answer

ok its ok then :)