State Cauchy’s criterion and Weierstrass’s test for uniform convergence for series. Test the uniform convergence of the series `\sum u_n ; u_n = \frac{e^ -nx}{4n^2 - 1}` by using Weierstrass’s test.

State Cauchy’s criterion and Weierstrass’s test for uniform convergence for series. Test the uniform convergence of the series `\sum u_n ; u_n = \frac{e^ -nx}{4n^2 - 1}` by using Weierstrass’s test.

Solution:

Cauchy's Criterion for Uniform Convergence:

For a series of functions `\sum u_n (x)` to converge uniformly on a domain D, it is necessary that for every ϵ >0, there exists an N such that for all m,n ≥ N and for all x ∈ D, the inequality `\|u_n (x) +  u_{n + 1} (x) +  \ldots  + u_m (x)| < \in` holds.

Weierstrass’s M-Test for Uniform Convergence:

Suppose `\sum u_n (x)` is a series of functions defined on a set D and `\sum M_n` is a series of positive real numbers such that `| u_n (x)| \le M_n` for all x ∈ D and all n ≥ 1. If the series `\sum M_n` converges, then the series of functions `\sum u_n (x)` converges uniformly on D.

Now, let's test the uniform convergence of the series `\sum u_n` where `u_n = \frac{e^{-nx}}{4n^2 - 1}` using Weierstrass's M-Test.

We need to find Mn such that `|u_n (x)| \le M_n` for all x in a given domain.

Given `u_n = \frac{e^{-nx}}{4n^2 - 1}` , it's necessary to find an upper bound `M_n` that satisfies `| u_n (x) | \le M_n` for all x in the domain.

Notice that the exponential term `e^{-nx}` can be maximized by minimizing the exponent −nx. This occurs when x is at its maximum value. Considering x ≥ 0, as x increases, the value of `e^- nx` decreases.

So, let's find the maximum value of `\frac{e^-nx}{4n^2 - 1}` for x ≥ 0.

As x increases, the term `\frac{e^-nx}{4n^2 - 1}` decreases, as the denominator is increasing and the exponential term is decreasing.

To find an upper bound for `M_n`, consider the term at the point where it's minimized, at x = 0:

`M_n = \frac{e^0}{4n^2 - 1} = \frac{1}{4n^2 - 1}` 

Now, let's check if the series `\sum \frac{1}{4n^2 - 1}` converges. If it does, according to Weierstrass's M-Test, the series `\sum u_n` will uniformly converge.

To test the convergence of the series `\sum \frac{1}{4n^2 - 1}`, we need to use some convergence test methods such as the comparison test or the limit comparison test.

If the series `\sum \frac{1}{4n^2 - 1}` converges, then by Weierstrass's M-Test, the series  `\sum u_n`.  will uniformly converge on the specified domain.

This procedure doesn't directly apply Cauchy's Criterion but instead uses Weierstrass's M-Test to establish uniform convergence.