Question

How do you use the limit comparison test on this particular series?Calculus series tests​

Answer 1

Compare $\dfrac1{\sqrt{n^2+1}}$ to $\dfrac1{\sqrt{n^2}}=\dfrac1n$. Then in applying the LCT, we have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1$

Because this limit is finite, both

$\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}$

and

$\displaystyle\sum_{n=1}^\infty\frac1n$

behave the same way. The second series diverges, so

$\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n$

is divergent.