Solution

Solution

Solution steps
Apply the constant multiplication rule:
Apply Series Ratio Test:converges

Popular Examples

sum from k=2 to infinity of (1/3)^{3k}sum from n=1 to infinity of (0)^{n-1}sum from k=2 to infinity of (-0.9)^ksum from k=0 to infinity of (5/8)^ksum from n=5 to infinity of cos(1/(n+1))

Frequently Asked Questions (FAQ)

  • What is the sum from k=2 to infinity of 9e^{-2k} ?

    The sum from k=2 to infinity of 9e^{-2k} is converges