Solution

Solution

Solution steps
Apply the constant multiplication rule:
Apply Series Limit Comparison Test:diverges

Popular Examples

sum from n=1 to infinity of 8-6(0.8)^nsum from n=0 to infinity of 1/(2n^2+n)sum from n=2 to infinity of ne^{-5n^2}sum from n=-3 to infinity of 1/(3^{n+1)}sum from n=3 to infinity of ne^{-n^2}

Frequently Asked Questions (FAQ)

  • What is the sum from n=1 to infinity of (5n)/(n^2+1) ?

    The sum from n=1 to infinity of (5n)/(n^2+1) is diverges