Confused on what a series means.

With respect to the original formula,and

usually $v<<c$ $$E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}=\\mc^2(1-\frac{v^2}{c^2})^{-\frac12}\\ \sim mc^2 (1-(-\frac12)\frac{v^2}{c^2}+...)\\=mc^2+\frac 12mv^2+...$$

As the other answer is rather formal, let me provide my own short one, dealing with the problem at hand.

The exact expression for the relativistic kinetic energy is:

$$K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$$

That is, total energy minus the rest energy.

In the non-relativistic limit we have the following assumption:

$$v^2 \ll c^2$$

Or:

$$\beta=\frac{v^2}{c^2} \ll 1$$

Turns out that there's a useful sequence of approximations that converge to the square root, provided that $\beta$ is small in absolute value as indicated above.

$$\sqrt{1+\beta}=1+\frac{\beta}{2}-\frac{\beta^2}{8}+\cdots$$

Where $\cdots$ should be replaced by as many terms as it takes to get the desired accuracy.

This is a case of so called Taylor series, and I encourage the OP to look it up.

In this particular case, it's enough to take only the first term, which gives us:

$$K \approx mc^2 \left( \frac{1}{1-v^2/(2c^2)}-1 \right)$$

But this is still too complicated.

However, we know another Taylor series, or rather a simple case of geometric progression sum:

$$\frac{1}{1-\beta} = 1+\beta+\beta^2+\cdots$$

Again, it's enough to use the first term, which gives us:

$$K \approx mc^2 \cdot \frac{v^2}{2c^2}=\frac{mv^2}{2}$$

For what it's worth, if you don't want to get too formal, a convergent series is a sequence of sums with increasing number of terms that gives a good approximation to some finite value. Usually, we can even estimate the error of the approximation.

Example:

$$\frac{\pi}{4}=1-\frac13+\frac15-\frac17+\cdots$$

• the famous Leibniz series for Pi.

A good estimation for the error is half the last term.