The conditions of the comparison sum state that if $0\le a_n\le b_n$
- and if $b_n$ converges, then $a_n$ also converges
- and if $a_n$ diverges, then $b_n$ also diverges.
I'm not sure how to go about this question though - do I try and show that it is greater than $1/n$ and so diverges?
$\endgroup$ 35 Answers
$\begingroup$Hint: I assume that we are starting at $n=1$. Show instead that your terms are $\ge \dfrac{1}{3n}$. This is not hard, since $2\le 2n^3$.
Showing that the $n$-th term is $\ge \dfrac{1}{3n}$ is plenty good enough to show divergence, and uses only crude inequalities.
$\endgroup$ 5 $\begingroup$$$\frac{n^2+1}{n^3+2}\geq\frac{n^2}{2n^3}=\frac{1}{2n}\Longrightarrow \sum_{n=1}^\infty\frac{n^2+1}{n^3+2}\,\,\,\text{diverges}$$
$\endgroup$ $\begingroup$$\frac{n^2+1}{n^3 + 2} > \frac{n^2+1}{n^3 +n}$
for $n>2$
$\frac{n^2+1}{n^3+2} > \frac{n^2+1}{n(n^2+1)} = \frac{1}{n}$
$\frac{1}{n}$ is the harmonic series and is divergent. Hence said function is divergent.
$\endgroup$ $\begingroup$In fact, it is eventually greater than $\frac 1n$ as you can see by dividing through by $n^2+1$, getting $\frac 1{n-\frac 1n+\frac 2{n^2+1}}$. But it might be easier to prove it is greater than $\frac 1{2n}$, which also diverges.
$\endgroup$ $\begingroup$Note that if $0\le 2a\le b$ then $\cfrac {a+1}{b+2}\ge\cfrac ab$, since we have that for $n\ge 2, \ n^3\ge 2n^2$, then $$\sum_{n\ge 0}\cfrac{n^2+1}{n^3+2} =\cfrac 12+\cfrac 23 +\sum_{n\ge 2}\cfrac{n^2+1}{n^3+2} \ge \cfrac 76 +\sum_{n\ge 2}\cfrac{n^2}{n^3} = \cfrac 76 +\sum_{n\ge 2}\cfrac 1n \quad\text{ which diverges}$$
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