Solution

Solution

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

Popular Examples

sum from n=1 to infinity of n(-5/9)^nsum from n=0 to infinity of 1/(n^{1.5)}sum from n=1 to infinity of 9/(n^2+49)sum from n=0 to infinity of (-1)^nn^2sum from n=1 to infinity of ((-e)/pi)^n

Frequently Asked Questions (FAQ)

  • What is the sum from n=2 to infinity of 6/(nln(n)) ?

    The sum from n=2 to infinity of 6/(nln(n)) is diverges