Compare [tex]\dfrac1{\sqrt{n^2+1}}[/tex] to [tex]\dfrac1{\sqrt{n^2}}=\dfrac1n[/tex]. Then in applying the LCT, we have
[tex]\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[/tex]
Because this limit is finite, both
[tex]\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}[/tex]
and
[tex]\displaystyle\sum_{n=1}^\infty\frac1n[/tex]
behave the same way. The second series diverges, so
[tex]\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n[/tex]
is divergent.
