Solution
Solution
Solution steps
Apply the constant multiplication rule:
Apply Series Ratio Test:converges
Popular Examples
sum from k=0 to infinity of (-1/4)^ksum from n=1 to infinity of 6^nsum from n=0 to infinity of 7/nsum from n=1 to infinity of n!(-e)^{-6n}sum from n=0 to infinity}2^{2-n of*3^n
Frequently Asked Questions (FAQ)
What is the sum from n=5 to infinity of 3/(2^n) ?
The sum from n=5 to infinity of 3/(2^n) is converges