infinity of the summation n=1: (e^n)/(n!) [using the ratio test]
my work so far:
= lim (n->infinity) | [(e^n+1)/((n+1)!)] / [(e^n)/(n!)] |
= lim (n->infinity) | [(e^n+1)/((n+1)!)] * [(n!)/(e^n)] |
= lim (n->infinity) | ((e^n)(e^1)(n!)) / ((n+1)(n!)(e^n)) |
..the e^n & n! cancels out
= lim (n->infinity) | (e^1) / (n+1) |
im stuck here.. how do i finish this?
and also to find out if it's Divergent (L>1), convergent (L<1), or fails (L=1)?
For Further Reading
* Calculus - ratio test - Count Iblis, Sunday, July 29, 2007 at 7:01pm
The limit is zero. So, L = 0 therefore the series is convergent.
The value of the summation is e^e - 1
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thank you for your response, but..
how did you get the summation of e^(e-1)
...from this?
= lim (n->infinity) | (e^1) / (n+1) |
i think i figured it out.
0 answers