Solution
Solution
Solution steps
Apply the constant multiplication rule:
Apply Series Divergence Test:diverges
Popular Examples
sum from n=1 to infinity}(e^{1/n of-1)/nsum from n=1 to infinity of 4-9(0.6)^nsum from n=2 to infinity of 6/(n(n-1))sum from n=0 to infinity of (e)^{-n}sum from n=0 to infinity of (4/25)^n
Frequently Asked Questions (FAQ)
What is the sum from n=0 to infinity of 2nln(1+n) ?
The sum from n=0 to infinity of 2nln(1+n) is diverges