The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges. $$\sum_{n}^{\infty}(-1)^{n} \frac{2^{n}}{n !}\left(x^{2}+1\right)^{2 n}$$

Understanding the convergence of a series is crucial when dealing with power series.
Convergence is about whether the sum of the series approaches a finite value as more terms are added.
A power series takes the general form:
\(\sum_{n=0}^{\infty} a_n x^n\).
For our specific case, after substituting the variable and simplifying, we got:
\(\sum_{n=0}^{\infty} (-1)^n \frac{2^n}{n!}u^n\).
This is a power series in terms of \(u = (x^2 + 1)^2\).
Power series will converge within a certain range of 'x' values.
For the exponential series form \(e^x\), the series converges for all real numbers.
Here, since we are dealing with \(e^{-2u}\) and \(u = (x^2 + 1)^2\), this series will also converge for all real 'x' values.
This is due to the non-negative nature of our transformation of 'u'.

The exponential series is one of the most important series in mathematics.
It is given by:
\( \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x.\)
This series converges for all real numbers 'x'.
In our exercise, after transforming the original series using the variable substitution, we get the series in an exponential form:
\(-e^{-2u}\).
This still represents a modified exponential series, hence, converging for all values of 'u'.
Remembering the properties of exponential functions helps us quickly determine convergence.
They are continuous and differentiable for all real numbers and their series form converges everywhere.

Substitution simplifies complicated expressions into more manageable forms.
In this exercise, we used \(u = (x^2 + 1)^2\).
The substitution transforms the original expression into a power series in terms of 'u'.
This is a powerful method since it takes a potentially complex-looking problem and places it into a familiar and easier-to-handle series.
After substitution, recognizing the new series form as an exponential series \(-e^{-2u}\), enables us to apply known properties and convergence criteria easily.
Substitution is a key tool in mathematical problem-solving that simplifies processes and leads to clearer insights.
Make sure you understand how to choose the right substitution and keep track of the relationship between original and substituted variables.