Does convergent p-series equal $\frac{1}{p-1}$ ?? For p-series, I am looking at a derivation for integral test for the general case of $\frac{1}{x^p}$.  
$$\int_{1}^{\infty}\frac{1}{x^p}dx$$
For $p>1$, I have the integral equaling $\frac{1}{p-1}$.  What does this value mean?  It just means the integral has a value, meaning the series is also convergent/finite.  But, it does not mean the series itself EQUALS $\frac{1}{p-1}$  Is that correct?   We do not know what the $p$-series adds up to, correct?
 A: Question 1
"I am looking for the derivation of integral test for general case $f(x)=\frac{1}{x^p}$"

See the statement of integral test is if $f(n)$ is monotone decreasing in the interval $[N,\infty)$ then $\sum_{n=N}^{\infty}f(n)$ converges and diverges according to $\int_{N}^{\infty} f(x) dx$ i.e. if the integral is finite then sum converges else diverges.
Say if $f(x)=\frac{1}{x^p}$ then the sum $\sum_{n=1}^{\infty}\frac{1}{x^p}$ converges if and only if
$\int_{1}^{\infty} x^{-p} dx$ converges.
And $\int x^{-p} dx=\frac{x^{-p+1}}{-p+1}$ now you see if $p>1$ then $1-p<0$ then$\lim_{x \rightarrow \infty} x^{1-p}=0$
but when $p \leq 1$ then
$\lim_{x \rightarrow \infty} x^{1-p}\rightarrow \infty$
 Hence the integral tends to infinity.
Question 2
"what does $\frac{1}{1-p}$ mean ?

It does not mean that sum converges to that rather it states about the "finite-ness" of integral which is required to conclude the convergence.

Question  3
"We does not know to what value it adds up. Correct ?"

If the sum never approaches to anything(diverging) then the answer is trivial. But if the sum is convergent then you can find using some techniques in complex analysis,this usually emphacised on a course on "analytic number theory". As far your 'general case' is concerned then it is a very famous one.
A: You are correct, e.g. for $p = 2$ we have
$$\sum\limits_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \neq 1 = \int_1^{\infty} \frac{1}{x^2} \, \mathrm{d}x.$$
