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

Question

asked Jul 23, 2020 176k views

1 Answer

Stuart Herring · Answer 1 · 2020-07-30T02:50:58+0000

Compare
$\frac1{√(n^2+1)}$ to
$\frac1{√(n^2)}=\frac1n$ . Then in applying the LCT, we have

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

Because this limit is finite, both

$\displaystyle\sum_(n=1)^\infty\frac1{√(n^2+1)}$

and

$\displaystyle\sum_(n=1)^\infty\frac1n$

behave the same way. The second series diverges, so

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

is divergent.