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.
To finish evaluating the limit, you can simplify the expression as follows:
= lim (n->infinity) (e^1) / (n+1)
Since e^1 is a constant, you can take it out of the limit:
= e / lim (n->infinity) (n+1)
As n approaches infinity, the denominator (n+1) also approaches infinity. Therefore, the limit is infinity divided by infinity, which is an indeterminate form.
To resolve this, you can use L'Hopital's rule, which states that if the limit of a fraction f(x) / g(x) is of the indeterminate form 0/0 or infinity/infinity, then the limit of the derivative of f(x) divided by the derivative of g(x) is the same:
= e / lim (n->infinity) (1)
Since the denominator is now a constant (1), the limit simplifies to:
= e
Therefore, the sum of the series is e. Hence, the series converges to e, and the value of the summation is e.
To finish evaluating the limit, you just need to take the limit as n approaches infinity of (e^1)/(n+1).
Since e^1 is a constant, it remains the same as n approaches infinity. So, we can write the limit as:
lim (n->infinity) [(e^1)/(n+1)]
Now, as n approaches infinity, the denominator (n+1) becomes much larger than the numerator (e^1), effectively making the fraction approach zero. Therefore, the limit is 0.
So, you correctly found that the limit is zero. This implies that the series is convergent, according to the ratio test.
As for the value of the summation, you can find it by substituting n=infinity into the series:
∑ (e^n)/(n!) = (e^1)/(1!) + (e^2)/(2!) + (e^3)/(3!) + ...
Since the series is convergent, we can calculate its value. For this specific series, the sum is e^e - 1.
Therefore, the value of the given series is e^e - 1.