sigma k=2 to infinity, 1/k^2-1
how do you break up the k^2-1 into a minus part? i cant do it theres no partial fraction form i can ffind? please help with a little bit
Is...
$\displaystyle \displaystyle \frac{1}{k^{2}-1} = \frac{1}{(k-1)\ (k+1)} = \frac{1}{2} \ (\frac{1}{k-1} - \frac{1}{k+1})$ (1)
... so that...
$\displaystyle \displaystyle \sum_{k=2}^{\infty} = \frac{1}{2} (1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + ...) = \frac{1}{2}\ (1+\frac{1}{2}) = \frac{3}{4}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
how do you work out this part? how would you make that jump because i'm fine with the difference of two squares part but i couldnt reach the next part myself. the reason i ask how is because i dont wanna have to ask for the help for each question if i know the logic i can do my remaining questions just fine
thank you
im stuck on another one:
it is sigma k=2 to infinity again, but the sum is 1/k^3-k
the reason i am stuck is the partial fracs give three parts then i dont know what to do
i solved it as -1/k + (1/2)/k+1 + (1/2)/k-1, i am lost because that doesnt really help me at all
This is the rules of this forum.
If you post new problems in the same thread, It will be too hard to follow the replies.
Anyway, as a student taking infinite series, you should be able to find the partial fraction expansion.
$\displaystyle \dfrac{1}{k^3-k}=\dfrac{1}{k(k^2-1)}=\dfrac{1}{k(k-1)(k+1)}=\dfrac{A}{k}+\dfrac{B}{k-1}+\dfrac{C}{k+1}$
You do not know how to find A,B & C ?