Mentor
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Looks good!
Now, since n2 + 1/n > n2, for all n >= 1,arl146 said:ok well
we need [itex]\frac{n}{n^3+1}[/itex] < [itex]\frac{1}{n^2}[/itex] for the test
and that can be proven by:
[itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{n(1)}{n(n^2+1/n)}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex]
arl146 said:so now we are looking at [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] for my series to be convergent since [itex]\frac{1}{n^2}[/itex] converges.
looking at the denominators: n2+[itex]\frac{1}{n}[/itex] is > n2
thus making [itex]\frac{n}{n^3+1}[/itex] = [itex]\frac{1}{n^2+1/n}[/itex] < [itex]\frac{1}{n^2}[/itex] this true
is that good enough?