Help with simple math question

[Mike_B]

(\sum_{m=1}^{\infty} \frac{1}{m(m+1)(m+2)}=\frac{1}{4})
What are the steps involved in solving the above?
I broke into partial fractions : (1/2m) - (1/m+1) + (1/2m+4)
However, that does not seem to converge to 1/4.
What am I doing wrong?

 

[IlyaKEightSix]

Solving for what? It already tells you what m is, which is {1...infinity}. And from simply doing a few iterations on Google, it does seem to be converging to 1/4.

 

[Mike_B]

I apologize for asking the question in a confusing manner. I was attempting to solve the infinite sum using partial fractions and could not do so. I looked up the solution (1/4) on Wolfram, but still cannot get to it using the partial fractions.

 

[Li Cai]

You are already half way there.
sum 0f (1/2m) = 1/2 + 1/4 + 1/6 +1/8 ...
sum of (1/m+1)= 1/2+ 1/3 +1/4 +1/5 ...
sum of (1/2m+4) =1/6 +1/8 +1/10...

Add sum 1 and sum 3 you get 1/2 + 1/4 +2(1/6+1/8...)=1/2 + 1/4 +(1/3 +1/4 +1/5...)
Then subtract sum2 you get 1/2 + 1/4 + (1/3 + 1/4 + 1/5...)-1/2 -(1/3 + 1/4 + 1/5...) = 1/4

 

[Mike_B]

You are already half way there.
sum 0f (1/2m) = 1/2 + 1/4 + 1/6 +1/8 ...
sum of (1/m+1)= 1/2+ 1/3 +1/4 +1/5 ...
sum of (1/2m+4) =1/6 +1/8 +1/10...

Add sum 1 and sum 3 you get 1/2 + 1/4 +2(1/6+1/8...)=1/2 + 1/4 +(1/3 +1/4 +1/5...)
Then subtract sum2 you get 1/2 + 1/4 + (1/3 + 1/4 + 1/5...)-1/2 -(1/3 + 1/4 + 1/5...) = 1/4

I see it now. Thanks Li Cai.

 

[euroazn]

You are already half way there.
sum 0f (1/2m) = 1/2 + 1/4 + 1/6 +1/8 ...
sum of (1/m+1)= 1/2+ 1/3 +1/4 +1/5 ...
sum of (1/2m+4) =1/6 +1/8 +1/10...

Add sum 1 and sum 3 you get 1/2 + 1/4 +2(1/6+1/8...)=1/2 + 1/4 +(1/3 +1/4 +1/5...)
Then subtract sum2 you get 1/2 + 1/4 + (1/3 + 1/4 + 1/5...)-1/2 -(1/3 + 1/4 + 1/5...) = 1/4

Although this solution is pretty specious... note how each of those sums is infinite!

 

[Li Cai]

Although this solution is pretty specious... note how each of those sums is infinite!

Hmm why does it matter that the sums are infinite?

 

[euroazn]

Hmm why does it matter that the sums are infinite?

It's fine in this case as long as you do all of your arithmetic inside the sigma but you shouldn't be adding/subtracting infinite sums since re-ordering them can actually change convergence properties.

 

[pruse]

1/2 - 1/2 + 1/6
1/4 - 1/3 + 1/8
1/6-1/4+1/10
1/8-1/5+1/12
1/10-1/6+1/14
...

Notice how terms cancel diagonally: (1/6+1/6-1/3), (1/8+1/8-1/4), ... the only term that survives is 1/4.

 

[lordvid]

(\sum_{m=1}^{\infty} \frac{1}{2m(m+1)(2m+4)}=\frac{1}{4})
What are the steps involved in solving the above?
I broke into partial fractions : (1/2m) - (1/m+1) + (1/2m+4)
However, that does not seem to converge to 1/4.
What am I doing wrong?

Are you asking to prove this identity?

If so, I think there may be a typo....the infinite series you have written down on the left hand side of that identity sums to 1/16...not 1/4 (you can check with MATLAB).

The partial fraction sum you have written down sums to 4 times the fraction that is being infinitely summed.

So you have 1/4 in front of the proof that others have given...

Also, the comment about regrouping the infinite series changing the answer is nonsense, coz nowhere are we dealing with an alternating series...each infinite series we are considering is non-alternating...each partial sum in the series is monotonically increasing.

 

[Andy Nguyen]

The initial series was linked to a picture on Wolfram but it did not work as of yesterday so I latex it in from memory and it may be not the same series anymore.
Anyone remember what was the original series and I can correct them.

 

[Mike_B]

It was 1/(m(m+1)(m+2))

 

[Andy Nguyen]

Ok, the series in OP is corrected. Next time, just use Latex instead of linking to some websites which may break on a wimp.

 

[foudas]

If you take the partial sum
(\sum_{m=1}^{N}\frac{1}{m(m+1)(m+2)}\right))
for some (N>0) then after deleting diagonally the remaining terms you get partial sum equal to
(\frac{1}{2}\large(\frac{1}{N+2}-\frac{1}{N+1}+\frac{1}{2}\right))
which has limit 1/4 when (N \to \infty).
Sorry if I am wrong I didn't really check it.