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)?

The limit is zero. So, L = 0 therefore the series is convergent.

The value of the summation is e^e - 1

To finish solving this using the ratio test, you have correctly simplified the expression to:

lim (n -> infinity) | (e^1) / (n+1) |

Now, let's evaluate this limit. As n approaches infinity, the term (n+1) in the denominator becomes significantly larger than 1. Therefore, we can treat it as a large constant and ignore it in the limit calculation.

lim (n -> infinity) | (e^1) / (n+1) |

= (e^1) * lim (n -> infinity) 1 / (n+1)

Since the limit of 1/(n+1) as n approaches infinity is 0, we have:

(e^1) * 0 = 0

Therefore, the limit of the ratio is 0. Since the limit is less than 1, the series converges.

To find the value of the summation, you can use the formula for the sum of an infinite geometric series:

S = a / (1 - r)

Where a is the first term of the series and r is the common ratio. In this case, a = e^1 and r = 1/(n+1).

S = (e^1) / (1 - 1/(n+1))

Now, take the limit of S as n approaches infinity:

lim (n -> infinity) [(e^1) / (1 - 1/(n+1))]

Plugging in infinity for n, we get:

lim (n -> infinity) [(e^1) / (1 - 1/infinity)]

Since 1/infinity is 0, we have:

lim (n -> infinity) [(e^1) / (1 - 0)]

= e

Therefore, the value of the summation is e.

So, the series is convergent with a value of e.