Question

How do you use the limit comparison test to determine if `Sigma 1/(nsqrt(n^2+1))` from `[1,oo)` is convergent or divergent?

Answer 1

By comparing to `sum_(n=1)^oo1/n^2`, we find that `sum_(n=1)^oo1/(nsqrt(n^2+1))` is convergent through the limit comparison test.

Explanation:

The limit comparison test states that if `L=lim_(nrarroo)a_n/b_n` where `L` is a positive, finite value, then `suma_n` and `sumb_n` either both converge or both diverge.

When working with the limit comparison test, it's often helpful to let the series we are working with be `suma_n`. Thus, let `a_n=1/(nsqrt(n^2+1))`.

The trickier thing to do is choose `b_n`. The best thing to do is make `b_n` mimic whatever `a_n` does at infinity.

As `1/(nsqrt(n^2+1))` gets larger and larger, the constant becomes more and more irrelevant and we see that `a_n~~1/(nsqrt(n^2))=1/(n(n))=1/n^2`.

So, let `b_n=1/n^2`.

Now going to the limit comparison test, let `L=lim_(nrarroo)a_n/b_n`.

Then `L=lim_(nrarroo)(1/(nsqrt(n^2+1)))/(1/n^2)=lim_(nrarroo)n^2/(nsqrt(n^2+1))`

`=lim_(nrarroo)n/sqrt(n^2+1)`

`=lim_(nrarroo)n/(nsqrt(1+1/n^2)`

`=lim_(nrarroo)1/sqrt(1+1/n^2`

`=1/sqrt(1+0)`

`=1`

So `L=1`. Since this is positive and finite, we know that `suma_n` and `sumb_n` have the same convergence or divergence.

We know that `sumb_n=sum_(n=1)^oo1/n^2` converges through the p-series test, so this means that `suma_n=sum_(n=1)^oo1/(nsqrt(n^2+1))` also converges through the limit comparison test.