Solution
Solution
Solution steps
Apply the constant multiplication rule:
Apply Series Comparison Test:converges
Popular Examples
sum from n=0 to infinity of (n^2)/(4^n)sum from n=1 to infinity of 5/(6^{n-1)}sum from n=0 to infinity of n/(10^n)sum from n=2 to infinity of 1/((n^2-1))sum from n=1 to infinity of n/(n+4)
Frequently Asked Questions (FAQ)
What is the sum from n=0 to infinity of 4/(n^2+1) ?
The sum from n=0 to infinity of 4/(n^2+1) is converges