Question

How do you use the limit comparison test for `sum ((sqrt(n+1)) / (n^2 + 1))` as n goes to infinity?

Answer 1

Remembering that:

`sum_(n=0)^(+oo)1/n^alpha` is convergent if `alpha>1`, than:

`(sqrt(n+1)) / (n^2 + 1)~sqrt(n)/n^2=n^(1/2)/n^2=1/n^(3/2)`,

so that series is convergent.

Answer 2

This series behaves like `sqrtn/n^2 = 1/n^(3/2)`

Compare it to that using the limit comparison test:

`(sqrt(n+1)/(n^2+1))/(1/(n^(3/2))) = (sqrt(n+1)/(n^2+1)) n^(3/2)/1=(n^(3/2)sqrt(n+1))/(n^2+1)`

`= (n^(3/2)sqrtn sqrt(1+1/n))/(n^2(1+1/n^2)) = (sqrt(1+1/n))/(1+1/n^2)`

The limit as `n rarr oo` is 1, which is positive.

Since `sum 1/n^(3/2)` converges (p-series), so does `sum(sqrt(n+1)/(n^2+1)`