Question

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

Answer 1

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.