Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

5 Answers 5

up vote 1 down vote accepted

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.

share|improve this answer
    
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

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

share|improve this answer

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

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.

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.