Question

Use the Nth term divergence test to show the series (sin(n)/(n^3 +1) is inconclusive

Answer 1

`\displaystyle\sum_(n\ge0)(\sin n)/(n^3+1)`

You have


`-\frac1{n^3+1}\le(\sin n)/(n^3+1)\le\frac1{n^3+1}`

with
`\pm\frac1{n^3+1}\to0` as
`n\to\infty`, so by the squeeze theorem,


`\displaystyle\lim_(n\to\infty)(\sin n)/(n^3+1)=0`

and so the
`n`th term test fails/doesn't confirm that the series diverges.

It does, however, converge by the comparison test. Consider the convergent


`\displaystyle\sum_(n\ge1)\frac1{n^3}`

Because this converges, the alternating series


`\displaystyle\sum_(n\ge1)((-1)^n)/(n^3)`

converges absolutely, and


`\left|(\sin n)/(n^3+1)\right|\le\frac1{n^3+1}<\frac1{n^3}`