Solution

Solution

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

Popular Examples

ratiotest sum from n=1 to infinity of nsum from n=0 to infinity of 1/(n3^n)sum from n=1 to infinity of 3/((2n-1))sum from k=0 to infinity of (-3/10)^ksum from n=1 to infinity of n(-e)^{-n}

Frequently Asked Questions (FAQ)

  • What is the sum from n=1 to infinity of (6n)/(e^n) ?

    The sum from n=1 to infinity of (6n)/(e^n) is converges