Is this square root approximation correct?

Question

Playing around I found a series which looks to converge to the square root function.

$$\sqrt{p^2+q}\overset{?}{=}p\left(1-\sum_{n=1}^{+\infty}\left(-\frac q{2p^2}\right)^n\right)$$

Is it correct?

Answer 1

No, it is not correct. Your right-hand expression is a geometric series for $$p\left(1-\frac{-\frac q{2p^2}}{1+\frac q{2p^2}}\right),$$ which is a rational expression in $p$ and $q$.

The correct answer is given by Newton's "generalized binomial theorem", in the form $$\sqrt{p^2+q}=p\sqrt{1+q/p^2} = p\sum_{k\ge 0} \binom{1/2} k \left(\frac q {p^2}\right)^k,$$ which converges for $|q/p^2|<1$. The first two values of $\binom{1/2} k$ are $1$ and $1/2$, which match what you have. But $\binom{1/2}{2}=-1/8$ which does not match your $-1/4$.

Answer 2

No.

$$p\left(1-\sum_{n=1}^{+\infty}\left(-\frac q{2p^2}\right)^n\right)=p\left(1+\frac q{2p^2}\frac1{1+\dfrac q{2p^2}}\right)=p+\frac{pq}{2p^2+q}\ne\sqrt{p^2+q}.$$