Math Help - Convergence of the series 1 / n(n+2)

1. Convergence of the series 1 / n(n+2)

Hi,

I'm just about to hit my head on a table or a wall...

I have to determine if the series (sum symbol) (n=1 to infinity) 1 / n(n+2) is convergent or not, and if it is, what is its limit. In the book, the answer is "3/4".

So I began writing its partial sum :
sn = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

I noticed that every term in sn is multiplied by : n(n+2) / (n+1)(n+2+1) = n(n+2) / (n+1)(n+3)

So I thought it was a geometric series with
a = 1 / a(n+2)
r = n(n+2) / (n+1)(n+3)

Since r will always be < 1 for every n >= 1, I tried to resolve a / (1 - r), but I never got 3/4.

Thinking I had to set a equal to "1/3", I did not get 3/4 either.

Then, I told myself it was a combinaison of two series :
1. 1 / n2
2. 1 / 2n

But the second one does not have a limit since it is 1 / 2 multiplied by the harmonic series (1 / n) which does not converge.

What am I doing wrong ? What did I not see ?

I just started this calculus class in university and my college maths are soooooooo far away (11 years).

3. Re: Convergence of the series 1 / n(n+2)

Originally Posted by NZAU1984
Hi,

I'm just about to hit my head on a table or a wall...

I have to determine if the series (sum symbol) (n=1 to infinity) 1 / n(n+2) is convergent or not, and if it is, what is its limit. In the book, the answer is "3/4".

So I began writing its partial sum :
sn = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

I noticed that every term in sn is multiplied by : n(n+2) / (n+1)(n+2+1) = n(n+2) / (n+1)(n+3)

So I thought it was a geometric series with
a = 1 / a(n+2)
r = n(n+2) / (n+1)(n+3)

Since r will always be < 1 for every n >= 1, I tried to resolve a / (1 - r), but I never got 3/4.

Thinking I had to set a equal to "1/3", I did not get 3/4 either.

Then, I told myself it was a combinaison of two series :
1. 1 / n2
2. 1 / 2n

But the second one does not have a limit since it is 1 / 2 multiplied by the harmonic series (1 / n) which does not converge.

What am I doing wrong ? What did I not see ?

I just started this calculus class in university and my college maths are soooooooo far away (11 years).

This is a telescopic series!

We can rewrite the summand as

$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n} + \frac{-1}{n+2} \right)$

The above can be found by partial fraction decomposition (or many other ways)

$\sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac{1}{2} \sum_{n=1}^{\infty}\left( \frac{1}{n}+\frac{-1}{n+2} \right)$

Notice if you write out a few terms you will see that pattern...

$\frac{1}{2} \left[ \left( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ ...\right)-\left( \frac{1}{3}+\frac{1}{4}+ ...\right) \right]$

$\frac{1}{2} (\left( 1+\frac{1}{2} \right)=\frac{3}{4}$

4. Re: Convergence of the series 1 / n(n+2)

I just forgot about simplification, I really don't see that stuff anymore (I used to be so good in maths).

Thanks !

Just a little question,

At the very end of sn, there would remain $\frac{-1}{n+2}$ which tends towards 0 as n tends to zero, right ? So that's why we only have left $1 \cdot (\frac{1}{1}+\frac{1}{2})$ ?

5. Re: Convergence of the series 1 / n(n+2)

Originally Posted by NZAU1984
Just a little question,
At the very end of sn, there would remain $\frac{-1}{n+2}$ which tends towards 0 as n tends to zero, right ? So that's why we only have left $1 \cdot (\frac{1}{1}+\frac{1}{2})$ ?
For $n\ge 3$, each partial sum collapses to $S_n = \frac{1}{2}\left[ {1 + \frac{1}{2} - \frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right]$

The sequence of partial sums converges $\left( {S_n } \right) \to \frac{1}{2}\left[ {1 + \frac{1}{2}} \right] = \frac{3}{4}$

6. Re: Convergence of the series 1 / n(n+2)

Originally Posted by NZAU1984
Hi,

I'm just about to hit my head on a table or a wall...

I have to determine if the series (sum symbol) (n=1 to infinity) 1 / n(n+2) is convergent or not, and if it is, what is its limit. In the book, the answer is "3/4".

So I began writing its partial sum :
sn = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

I noticed that every term in sn is multiplied by : n(n+2) / (n+1)(n+2+1) = n(n+2) / (n+1)(n+3)

So I thought it was a geometric series with
a = 1 / a(n+2)
r = n(n+2) / (n+1)(n+3)
That's not a geometric series. A Geometric series is of the form $\sum ar^n$ with a and r constants, not depending on n.

Since r will always be < 1 for every n >= 1, I tried to resolve a / (1 - r), but I never got 3/4.

Thinking I had to set a equal to "1/3", I did not get 3/4 either.

Then, I told myself it was a combinaison of two series :
1. 1 / n2
2. 1 / 2n

But the second one does not have a limit since it is 1 / 2 multiplied by the harmonic series (1 / n) which does not converge.

What am I doing wrong ? What did I not see ?

I just started this calculus class in university and my college maths are soooooooo far away (11 years).

7. Re: Convergence of the series 1 / n(n+2)

Originally Posted by TheEmptySet

$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n} + \frac{-1}{n+2} \right)$
This is the trick to solving this problem and others like it.

- Hollywood

8. Re: Convergence of the series 1 / n(n+2)

There is a general result:if $q$ is a positive integer and $\varphi (n)$ a convergent sequence, $S=\sum_{n=1}^{\infty}(\varphi (n+q)-\varphi(n))=q\lim_{n\to \infty}\varphi(n)-(\varphi(1)+\ldots +\varphi(q)).$ In our case, $\varphi(n)=\frac{-1/2}{n}$ and $q=2,$ so $S=-\varphi(1)-\varphi(2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}.$