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 :

s_{n} = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

I noticed that every term in s_{n} 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 / n^{2}

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).

Thanks for your help :)

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

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

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 which tends towards 0 as n tends to zero, right ? So that's why we only have left ?

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

Quote:

Originally Posted by

**NZAU1984** Just a little question,

At the very end of sn, there would remain

which tends towards 0 as n tends to zero, right ? So that's why we only have left

?

For , each partial sum collapses to

The sequence of partial sums converges

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

Quote:

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 :

s_{n} = 1/3 + 1/8 + 1/15 + 1/24 + ... + 1 / n(n+2) + ...

I noticed that every term in s_{n} 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 with a and r **constants**, not depending on n.

Quote:

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 / n^{2}

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).

Thanks for your help :)

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

Quote:

Originally Posted by

**TheEmptySet**

This is the trick to solving this problem and others like it.

- Hollywood

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

There is a general result:if is a positive integer and a convergent sequence, In our case, and so