Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.
I've tried to start this as follows: Assuming that m>n, we have
|a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.
So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!
Yes, that's what I meant...couldn't figure out how to do the Latex code here.
Okay, so I need to show that
.
But again I am faced with finding an N such that m,n>N implies that this is less than any given epsilon. But to do that, I should somehow eliminate m, yes?
Thanks again!
It is possible to prove this using an integral estimate.
Consider the function, defined on the positive reals. We will place a good bound on the quantity where . Consider the function on the interval . The area is equal to . Now consider the right-endpoint rectangular approximation of the area equally spaced at integer points between and (each sub-rectangle has width ) this approximation is clearly smaller then the area, furthermore its quantity is .
Thus, . Note, can be made arbitrary small if we pick so large that .