Determine if the series converges or diverges.
Sum (n=1, infinity) (2*4*6...2n)/n!
My main problem is I have no idea what to do with the numerator.
The numerator is 2^n * n!
That help?
divide by n! without 2n!
To determine if the series converges or diverges, we need to examine the behavior of the terms as n approaches infinity. Let's break down the terms:
The numerator is (2*4*6*...*2n), which is a product of consecutive even numbers. We can rewrite this as (2^n)(1*2*3*...*n) since 2^n is common to every term.
The denominator is n!, which represents the factorial of n.
Now, we have:
Sum (n=1, infinity) [(2^n)(1*2*3*...*n) / n!]
To further simplify the expression, we can rearrange the terms:
= Sum (n=1, infinity) [(2^n / n!) * (1*2*3*...*n)]
Now, let's focus on the ratio (2^n / n!). As n approaches infinity, 2^n increases much faster than n!. This is because the exponential growth of 2^n outpaces the factorial growth of n!. Hence, this ratio tends to infinity.
Since the ratio (2^n / n!) tends to infinity, the series diverges.