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!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
Hmm why does it matter that the sums are infinite?Although this solution is pretty specious... note how each of those sums is 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.Hmm why does it matter that the sums are infinite?
(\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?