what is the difference between sum of first n primes and prime(prime(n?

How does the difference method work?

• Hi, I have used the method of differences (also called a telescoping series) to solve various maths problems before but I don't feel I understand the mechanics of it. I began looking for a more general definition of it and found this in the Mathematical Methods for Physics and Engineering Textbook: \sum _{n=1}^N u_n=u_1+u_2+\text{...}+u_N with u_n being defined as:   u_n=f(n)-f(n-1) "for some function f[n]" we can see that everything in the series will cancel apart from the first and last value: (f(1)-f(0)) + (f(2)-f(1))+ ...+(f(N)-f(N-1)) = f(N)-f(0) leading to the general form: S_n=\sum _{n=1}^N u_n=f(N)-f(0) The method above is in contrast with one described by other sources that define: \sum _{n=1}^N u_{n+1}=u_2+u_3+\text{...}+u_{N+1} with u_n being defined as: u_n=f(n+1)-f(n) which leads to a general form of: S_n=\sum _{n=1}^N u_{n+1}=f(N+1)-f(1) for example both can be used to prove:   \sum _{n=1}^N n=\frac{1}{2} N (N+1) So I guess my question is what's the difference between these two forms, is there a way one can generalise the two into one? Thanks

• Answer:

  You must have miswritten the second method. If u_n = f(n + 1) - f(n), then \displaystyle \sum_{n = 1}^{N} u_{n + 1} = u_2 + u_3 + ... + u_{N + 1} will be equal to f(N + 2) - f(2), not f(N + 1) - f(1). At any rate, the differences you've indicated are entirely trivial; they're just "reindexing" (changing one's mind about whether the summation should start with f(1) - f(0), or f(2) - f(1), or whatever, and whether the starting term should be assigned a nominal index of 0, or 1, or 2, or whatever). If they really seem to you like different methods, you are bizarrely hung up on irrelevant, picayune details, rather than the key idea (that adding a bunch of consecutive differences yields a cumulative difference).