I'm having trouble understanding something regarding the p-series. (Post below in just a minute).

Question

mendicant_bias · Answer

What exactly is the value 1/p-1 in the case that p > 1 and the series converges? I know that's some calculable and relevant value, but it makes a point of stating it's not the sum of the p-series; what is it?

anonymous · Answer

it is not the sum of the series, it is just a test to see if the series converges or not
it does not give the actual value because the series is not equal to the integral
the integral computes the areas under a smooth curve, the series is a step function

mendicant_bias · Answer

There's another part of the book-and I think I just figured out why-that states that for a p-integral where p > 1, the improper integral* converges to the value of 1/p-1. I'm guessing this isn't valid in the first instance because it's essentially a comparison test? I don't know.

mendicant_bias · Answer

attachment

anonymous · Answer

If $a_n=f(n)$ then:$$
\int _1^\infty f(\lfloor x\rfloor )\;dx=\sum_1^\infty a_n
$$So we can use this as a basis for comparison if we know that $$
f(\lfloor x\rfloor )<f(x)
$$or $$
f(\lfloor x\rfloor ) > f(x)
$$

mendicant_bias · Answer

I'll probably sound like a jerk to just now say this, but I just got what satellite was saying, lol. But yeah, in the second .PNG, it's sort of like a direct comarison test whicdefinitely means that they won't converge to the same valuewhle in the first, it's just an evaluation of something that will always closely enough approximate the real tthingsince th upper limit of integral is infinity (PS sorryusingablet, hard to be accurate & quick)