I think i have my way to get out from this confusion sir..
I see that the author of that book derive the equation from ##\sum_{n=1}^N u_n## he took the summation from ##n=1##. So we can use the formula directly only when the summation begin from ##n=1##. So i want to generalized this result for...
I understand what you mean sir... But there is something bothering me from the book... For example : Evaluate the sum $$\sum_{n=1}^N \frac 1 {n(n+2)}$$
Using partial fraction we find $$u_n=-\left[\frac 1{2(n+2)}-\frac 1{2n}\right]$$
Hence ##u_n=f(n)-f(n-2)## with ##f(n)=-\frac 1 {2(n+2)}##, and...
That was my first approach to the problem sir.. But there is something bother me.. From that we get ##f(n)=-\frac 1 {2n^2}## and if we use the formula of ##S_N## we get ##S_N=f(N)-f(0)##. The problem is the function is undefined at ##n=0##. That's why i try to change the original problem.
I forgot to wrote it Sir.. Sorry, my bad... :')
##f(n)## is defined as a function of n such that ##u_n=f(n)-f(n-m)## with m as a 'difference' between the two (i mean like 5-3=2, 5 is differ from 2 by 3, so in this example i have 3 as my 'm'. Sorry i can't be more rigour like a mathemathician...
Yes Sir, i already derive it by myself. A little different, but it produce the same thing. This is what i got from my derivation ##\sum_{k=1}^m f(N-m+k) - \sum_{k=1}^m f(m-k)## like i wrote above.. I knew what m mean sir..
Homework Statement
Use the diffence method to sum the series ##\sum_{n=2}^N\frac {2n-1}{2n^2(n-1)^2}##
Homework Equations
from the textbook by Riley i have ##S_N=\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)##
The Attempt at a Solution
I prefer to write it as ##S_N=\sum_{k=1}^m...
Thank you Berkeman. May i ask a question? What Should i Study first to learn quantum mechanics by myself? The tuition fee in my country is so high, i can't afford it. But i still want to learn it badly.