Solution
Solution
Solution steps
Apply the constant multiplication rule:
Apply Series Ratio Test:converges
Popular Examples
sum from n=1 to infinity of (-1)^nsec(n)sum from n=1 to infinity of 1/(12n+11)sum from n=3 to infinity of (1/4)^nsum from n=1 to infinity of 1/((ln(n)))sum from n=0 to infinity of 3(2/6)^{n+2}
Frequently Asked Questions (FAQ)
What is the sum from n=1 to infinity of 7n^2e^{-n^3} ?
The sum from n=1 to infinity of 7n^2e^{-n^3} is converges