Proof that a series diverges

Question

How could I show that the series $$\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}$$ Diverges without the use of comparison test which I am able to show. Any hints would be great thanks.

Answer 1

To avoid the comparison test, compute $$\lim_{n\to \infty} \frac{n}{1}\frac{1}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1}{1+n^{-2}}=1$$

Hence, by the limit comparison test, the series agrees with $$\sum_{n\ge 1}\frac{1}{n}$$ which diverges.

Answer 2

Hint:With the integral test, we have that $$\int_0^{\infty} \frac{1}{\sqrt{x^2+1}}\mathrm{d}x$$ and $$\sum_{n=0}^{\infty} \frac{1}{\sqrt{n^2+1}}$$ are equiconvergent.

Answer 3

You can use the integral test:$$\int_0^\infty\frac{\mathrm dx}{\sqrt{x^2+1}}=\lim_{M\to\infty}\log\left(M+\sqrt{1+M^2}\right)=\infty.$$