How do you test the series #Sigma n/((n+1)(n^2+1))# from n is #[0,oo)# for convergence?

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Answer 1 · 2017-01-18T14:39:47

The series:

# sum_(n=0)^oo n/((n+1)(n^2+1))#

is convergent.

You can test it by direct comparison, considering that for #n>1#:

#0 < n/((n+1)(n^2+1)) < n/n^3 = 1/n^2#

As the series:

#sum_(n=0)^oo 1/n^2#

is convergent based on the p-series test, then also the series:

# sum_(n=0)^oo n/((n+1)(n^2+1))#

is convergent.