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Help with simple math question

Mike_B

Joined
8/15/11
Messages
62
Points
18
(\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?
 
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.
 
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.
 
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
 
Li Cai said:
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
Click to expand...

I see it now. Thanks Li Cai.
 
Li Cai said:
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
Click to expand...
Although this solution is pretty specious... note how each of those sums is infinite! ;)
 
Li Cai said:
Hmm why does it matter that the sums are infinite?
Click to expand...
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.
 
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.
 
Mike_B said:
(\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?
Click to expand...

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.
 
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.
 
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