The value of the driving frequency at which the voltage across the capacitor becomes maximum in a series RLC ac fed circuit The circuit diagram is as shown.
As per the book voltage across the capacitor is maximum during resonance that is p=1/√(LC)
But what I found out is a bit different .
And it is.
p=√((1/LC)-(R²/2L²))
I got this result by solving the 2nd order degree 1 differential equation in terms of q (charge) in capacitor and then found the voltage across across the capacitor as q/c and took its amplitude and differentiated it to get the result. I am not sure if I am right or wrong . please help me out 
 A: I solved this problem many years ago and scanned the solution into my computer (messy derivative). My result agrees with yours.  I also have a spreadsheet which solves RLC circuits (with numbers). I used “solver” to maximize the voltage on the capacitor, and found that the predicted angular frequency matched that suggested by your formula for several different sets of circuit parameters.
A: 
As per the book voltage across the capacitor is maximum during
  resonance that is p=1/√(LC)

is an incorrect statement, rather when that condition is satisfied you have current resonance and the current through and the voltage across the resistor is a maximum.
You are asked about charge resonance when the charge stored on the capacitor and the voltage across the capacitor is a maximum.
You have found that it occurs at a different frequency.
This is a related link.
A: 
"As per the book voltage across the capacitor is maximum during resonance that is p=1/√(LC)"

EDIT:  I looked closer after seeing R.W. Bird's answer. Updated below accordingly.  You can learn something every day if you pay attention.  ;--)
That is approximately correct statement for AC excitation when R is relatively small, but, apparently the OP equation is correct (and the exact solution).
Case 1:  I ran a frequency scan for a series RLC circuit with these parameters: R = 1Ω, L = 26.5mH, and C = 10μF. A plot of the capacitor voltage amplitude vs. frequency is shown below.  For a series circuit the resonance is a series resonance, |XC| = |XL|, so the current will be maximum at the resonant frequency.
$$ f  = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(26.5E-3)(10E-6)}}= 309Hz$$
$$ f  = \frac{\sqrt{\frac{1}{LC}-\frac{R^2}{2L^2}}}{2\pi}= 309Hz$$

Case 2:  I change R from 1Ω to 30Ω and re-ran. This time the peak voltage across the capacitor occured at 297Hz as per the OP's formula.
$$ f  = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{(26.5E-3)(10E-6)}}= 309Hz$$
$$ f  = \frac{\sqrt{\frac{1}{LC}-\frac{R^2}{2L^2}}}{2\pi}= 297Hz$$

Below is the same plot but i've added the voltage across the resistor too (red trace).  Interestingly, the peak voltage across the R occurs at 309Hz - which jives with the link that Farcher gives in his answer.

