The ratio f_n(x)/g_n(x) does NOT converge uniformly if at some x=x_0: g(x_0) = 0.

One of possible formulations to make the theorem CORRECT is the following:

"
There is a set X, there are sequences of functions f_n: X-->R and g_n: X-->R, both series converge uniformly, f_n-->f(x) and g_n-->g(x), and the limit functions have the following properties:
(a) both f and g are bounded, that is there are some F and G such that |f(x)| < F and |g(x)| < G for all x in X,
(b) g(x) has the property that there is some H>0 such that |g(x)| > H for all x in X.

Then f_n(x)/g_n(x) --> f(x)/g(x) uniformly.
"

This formulation can be modified.
Examples:
--- it is possible to replace the real numbers R by complex numbers C
--- it can be stated "if and only if".
--- condition (b) can be replaced by "the set X is closed and g(x) not = 0 in X"
+ there can be other modifications

Determine the uniformconvergence and convergence of the series ∑▒〖〖(f〗_n),〗 where f_n (x)is given by the following: (The Weoerstrass M-Test will be needed)
a sin(x/n^2 ) b. 〖(nx)〗^(-2),x≠0,
c. 〖(x^2+n^2)〗^(-1) d. (-1)^n (n+x)^(-1),x≥0,
e. 〖(x^n+1)〗^(-1),x≥0

1.Show that if a>0, then the convergence of the function nx/(1+n^2 x^2) is uniform on the interval [a,infinity), but is not uniform on the interval [0,infinity).
2. Show that if b is greater than 0 and less than 1, then the convergence of the function x^n/(1+x^n) is uniform on the interval [0,b] but is not uniform on the interv

1. Discuss the convergence and the uniformconvergence of the series sum (f_n) where f_n(x) is given by
a) x^n/(x^n+1) where x is greater than or equal to zero
b) ((-1)^n))/(n+x) for x greater than or equal to zero.

Prove the following theorem.
Let f1,f2,f3.... be continuous functions on a closed bounded interval [a,b] . Then fn--->f uniformly on [a,b] if and only if
fn(x)-->f(x) for every xn-->x such that xn,x E[a,b] .
Please see the attached file for the fully formatted problems.

Consider the sequence of functions f_n(x) =sin ([2npi)^2+x]^(1/2)) on [0,infinity)
a. Show that lim( n goes to infinity) f_n(x)=0
Hint: Use the mean value theorem for f_n(x) on [0,x]
b. Show that f_n(x) converges uniformly to 0 on [0,a] for a fixed a in [0,infinity).
c. Is it true that f_n(x) converges uniformly to

Prove that A sequence of functions (f_n) defined on a set A subset or equal to R converges uniformly on A if and only if for every e>0(epsilon) there exists an N belong to N such that Absolute value of f_n (x)-f_m (x)=N and all x belong to A.

Note: all of the following _n denote n as a subscript.
Suppose {f_n} from n = 0 to infinity, is a sequence of functions converging uniformly to a function f on an interval [a,b]. Also assume that the sequence has a uniform bound i.e. there exists M1 such that for all positive integers n, | f_n(x) | < = M1 for all x belonging

Let fk (x) = kxe - kx, k = 1, 2, 3, ...
It can be shown that the sequence {fk} infinity, k = 1, converges to 0
Pointwise on [0, +oo) but convergence is not uniform on [0, +oo).
a. Show that convergence is not uniform on [0, 1] either. Thus the sequence does not converge in the normed vector space (C ([0, 1]) , II·II in

(1) Let G = {z : 0 < abs(z) < R} for some R > 0 and let f be analytic on the punctured
disk G with Laurent Series f(z) = sum a_n*z^n (from n = -oo to oo).
(a) If f_n(z) = sum a_k*z^k (from k =-oo to n), then prove that f_n converges pointwise
f in C(G,C) (all continuous functions from G to C (complex)); i.e., {f_n}