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I tried posting this yesterday but I messed up typing the problem and I have now deleted that post (EDIT: oops, turns out you can't delete it after a day, sorry).

I'm trying to solve the problem in the title, but I'm having trouble justifying and making sure my answer is correct.

Also, I know that a problem like this has been solved before on here Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ but the answers on there make this more confusing since they get $1\over4$ as an answer and they don't seem to justify the answers they get.

So you know what experience I have and how I should try to justify my answer, here is a picture of the chapter and its sections.

enter image description here

I don't know which section the problem in the title comes from. This is also from a calculus 2 course.

Ok, first, if I take the limit of $$\sum_{n=1}^\infty {1\over n(n+1)(n+2)}$$ I will end up with $0$ correct? Since as n gets bigger the denominator also gets bigger meaning the number gets smaller. I'm not even sure if I need to take the limit. Assuming what I just did is correct, that means the series is convergent to $0$, but how would I justify the answer I get? If you look at the chapter I uploaded, is there a test I can use? I have tried reading all of it but it just isn't making sense to me.

Thank you for any help anyone can give me, If I can see how someone would go about solving this, then I should be able to solve the rest on my own.

Martin Sleziak
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JustHeavy
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    It compares with $\sum 1/n^3$. It also telescopes. – Angina Seng Oct 28 '17 at 17:40
  • I am sorry but the sum is a sum of positive terms the first of which is $1/6$. The terms themselves tend to zero, which is necessary, but not sufficient, for convergence. But the sum cannot possibly be zero, because it is greater than zero in its first term. You need to read the answers to the question you have linked and actually write the ideas they have down term by term to see how the best ones do work by cancellation. If you have a problem there, then come back and ask that question, but you need to do the work to understand what is going on here. – Mark Bennet Oct 28 '17 at 17:44
  • You seem to be confusing sequence for series. It's correct that the sequence $a_n=\frac1{n(n+1)(n+2)}$ goes to zero; in fact, that's a prerequisite for the series $\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac1{n(n+1)(n+2)}$ to converge. (But it's not sufficient - the harmonic series $\sum_n \frac1n$ is the classic example of a series where each individual term goes to zero but the sum doesn't converge.) – Steven Stadnicki Oct 28 '17 at 17:52

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$n(n+1)(n+2) >(n)(n+0)(n+0) \implies \sum \frac{1}{n(n+1)(n+2)} < \sum \frac{1}{n^3}$

By-p test as $p = 3 > 1$, given series converges

Magneto
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  • This is what I was looking for, so p is the exponent in the denominator, and by the p-test we know that it will be convergent. My only question is how do you get $1\over{n^3}$? Because if you multiply the denominator you get $n^3+3n^2+2n$. Do you just choose the term with the highest exponent? – JustHeavy Oct 28 '17 at 17:57
  • $n^3 + 3n^2 +2n > n^3$ inverse of above means inequality will have "less than" – Magneto Oct 28 '17 at 17:59
  • $a , b $ are positive and $a>b \implies \frac {1}{a} < \frac {1}{b}$ – Magneto Oct 28 '17 at 18:00