Solution
Solution
Solution steps
Apply the constant multiplication rule:
Apply Series Ratio Test:converges
Popular Examples
sum from n=1 to infinity of ne^{-4n^2}sum from n=1 to infinity of (n!)/(5+n)sum from n=5 to infinity of 1/(ln(n))sum from n=0 to infinity of n^{-1}sum from n=0 to infinity of 2-3(0.8)^n
Frequently Asked Questions (FAQ)
What is the sum from n=1 to infinity of 5n^2e^{-n^3} ?
The sum from n=1 to infinity of 5n^2e^{-n^3} is converges