Solution

Solution

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

Popular Examples

sum from n=1 to infinity of-1/(6^n)sum from k=4 to infinity of 3/(kln^2(k))sum from n=1 to infinity of (5n)/(n^2+1)sum from n=1 to infinity of 8-6(0.8)^nsum from n=0 to infinity of 1/(2n^2+n)

Frequently Asked Questions (FAQ)

  • What is the sum from n=0 to infinity of 7e^{-7n} ?

    The sum from n=0 to infinity of 7e^{-7n} is converges