# Using the Comparison Test, prove that the infinite series of $(n^2+1)/(n^3+2)$ converges/diverges.

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?

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you're on the right track. –  user31280 Nov 19 '12 at 0:48
@FOlaYinka I can't seem to show that the given series is greater than 1/n though. –  Mathlete Nov 19 '12 at 0:49
Equivalently: (n^2+1)/(n^3+2) - 1/n > 0. –  Charles Nov 19 '12 at 0:54

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.

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Yes, from n=1 to n=infinity, I'm still getting to grips with LaTex sorry. –  Mathlete Nov 19 '12 at 0:58
Would you start with the n-th term - 1/3n => 0 ? –  Mathlete Nov 19 '12 at 1:02
That does not help prove divergence. The point is that since $\sum\frac{1}{n}$ diverges, so does $\sum k\frac{1}{n}$ for any non-zero constant $k$. Here $k=\frac{1}{3}$. –  André Nicolas Nov 19 '12 at 1:04
I understand the principle of the comparison test, I'm just having trouble with the inequalities. How would I begin to show that the n-th term is greater than or equal to 1/3n? –  Mathlete Nov 19 '12 at 1:05
Numerator is $\gt n^2$. Denominator is $\le n^3+2n^3=3n^3$. Therefore ratio is $\gt \frac{n^2}{3n^3}$. –  André Nicolas Nov 19 '12 at 1:07

$$\frac{n^2+1}{n^3+2}\geq\frac{n^2}{2n^3}=\frac{1}{2n}\Longrightarrow \sum_{n=1}^\infty\frac{n^2+1}{n^3+2}\,\,\,\text{diverges}$$

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$\frac{n^2+1}{n^3 + 2} > \frac{n^2+1}{n^3 +n}$

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.

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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.
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 $$\sum_{n\ge 0}\cfrac{n^2+1}{n^3+2} =\cfrac 12+\cfrac 23 +\sum_{n\ge 2}\cfrac{n^2+1}{n^3+2} \ge \cfrac 76 +\sum_{n\ge 2}\cfrac{n^2}{n^3} = \cfrac 76 +\sum_{n\ge 2}\cfrac 1n \quad\text{ which diverges}$$