does the sum e^n / n! converge or diverge? n=1 --> infinity
To determine whether the sum of e^n / n! converges or diverges, we can use the ratio test.
The ratio test states that for a series ∑aₙ, if the limit of the absolute value of the ratio of consecutive terms, lim(n→∞) |aₙ₊₁ / aₙ|, is less than 1, then the series converges. If the limit is greater than 1 or it does not exist, then the series diverges.
Let's apply the ratio test to the given series. We have:
|aₙ₊₁ / aₙ| = |(e^(n+1) / (n+1)!)/(e^n/n!)|
= |(e^(n+1) / n+1) * (n! / e^n)|
= e * (n! / (n+1)!)
Simplifying further:
|aₙ₊₁ / aₙ| = e * (1 / (n+1)) (since n! / n! = 1)
Now, let's take the limit as n approaches infinity:
lim(n→∞) e * (1 / (n+1)) = e * 0 = 0
Since the limit is less than 1, we can conclude that the series of e^n / n! converges.