Convergence/Divergence of an Infinite Series

[Alex Myhill]

Homework Statement

To Determine Whether the series seen below is convergent or divergent.

Homework Equations

∑(n/((n+1)(n+2))) From n=1 to infinity.

The Attempt at a Solution

Tried to use the comparison test as the bottom is n^2 + 3n + 2, comparing to 1/n. However, this does not work as the first series is "smaller" than the harmonic series and thus it can't be proved to be divergent. I also tried to separate it into partial fractions but I don't know how to prove it is divergent from there. Thanks a lot for any help.

 

[fresh_42]

Homework Statement

To Determine Whether the series seen below is convergent or divergent.

Homework Equations

∑(n/((n+1)(n+2))) From n=1 to infinity.

The Attempt at a Solution

Tried to use the comparison test as the bottom is n^2 + 3n + 2, comparing to 1/n. However, this does not work as the first series is "smaller" than the harmonic series and thus it can't be proved to be divergent. I also tried to separate it into partial fractions but I don't know how to prove it is divergent from there. Thanks a lot for any help.

Is it still smaller if you start the harmonic series at six? And does this make the harmonic series convergent?

 

[Mark44]

Homework Statement

To Determine Whether the series seen below is convergent or divergent.

Homework Equations

∑(n/((n+1)(n+2))) From n=1 to infinity.

The Attempt at a Solution

Tried to use the comparison test as the bottom is n^2 + 3n + 2, comparing to 1/n. However, this does not work as the first series is "smaller" than the harmonic series and thus it can't be proved to be divergent. I also tried to separate it into partial fractions but I don't know how to prove it is divergent from there. Thanks a lot for any help.

Have you seen the limit comparison test?

Also, this is definitely not a precalculus problem, so I have moved it to the calculus section.

 

[Alex Myhill]

Hi,
Thanks for the response. So to prove using comparison just start harmonic series at six. Makes sense, I hadn't thought of that.
Referring to Mark44's response, no, I haven't heard of the limit comparison test, however, I will look it up.
Thanks heaps for your time and input.

 

[Ray Vickson]

Hi,
Thanks for the response. So to prove using comparison just start harmonic series at six. Makes sense, I hadn't thought of that.
Referring to Mark44's response, no, I haven't heard of the limit comparison test, however, I will look it up.
Thanks heaps for your time and input.

I don't see how starting the harmonic series at six solves the problem, unless you also do some other things at the same time.

To me, the simplest solution is to note that for $n \geq 1$ we have $n/(n+1) \geq 1/2$, so that
$$t_n \equiv \frac{n}{(n+1)(n+2)} \geq \frac{1}{2} \frac{1}{n+2}.$$
That means
$$\sum_{n=1}^N t_n \geq \sum_{n=1}^N \frac{1}{2} \frac{1}{n+2} = \frac{1}{2} \left[\frac{1}{3}+ \frac{1}{4} + \cdots + \frac{1}{N+2} \right] $$