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Determine the uniform convergence 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 f. x^n (x^n+1)^(-1),x≥0.

Suppose (k_n ) is a decreasing sequence of positive numbers. If ∑▒(k_n sin⁡nx ) is uniformly convergent, then lim⁡(nk_n )=0.

https://brainmass.com/math/real-analysis/uniform-convergence-proof-297769

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Determine the uniform convergence and convergence of the series where
a b.
c. d.
e. f.
a)
We know for all x, we have
For any given x, is convergent hence the series is convergent.
Further, as for any x. Hence the convergence is uniform.
b)
Let x is not in (-1, 1). Then we have . Hence the convergence of is uniform.
Let x is in (-1, 1) but not equal to zero. Let us suppose, . Then y is not in (-1, 1). So, we have
For a given such y, we have . Hence the convergence is pointwise.
Now, does not exist. Hence the convergence is not uniform in the interval (-1, 1)
c)
Let us first calculate the point wise limit for all x.
Now, . Hence the convergence is uniform for all x.

d)
Let us first calculate the point wise limit for all x.
Now, . Hence the convergence is uniform for all x .
e)
Let x is in [0, 1]. Then we know that
If x>1. Let . So, y is in (0, 1) and
As the convergence depends on the values of x, the convergence is not uniform
f)
Let x is in [0, 1]. Then we know that
If x>1. Let . So, y is in (0, 1) and
As the convergence depends on the values of x, the convergence is not uniform

Suppose is a decreasing sequence of positive numbers. If is uniformly convergent, then .

Given that is decreasing sequence of non-negative numbers and converges uniformly. So, corresponding to any , a positive integer m such that if n>m
Each term on the left hand side is greater than or equal to as is decreasing sequence.
Therefore, if n>m
Also, since converges. So, we choose N>m so that if n>N.
Thus for (n>m) gives for all n>N. Hence

Find the radius of convergence of the series where is given by
a. b.
c. d.

a. We have So, . This implies
Now consider, the ratio
Taking the limit as n tending to infinity for the absolute value of this ratio, we get as
The series converges if . Hence the radius of convergence is .

b. We have So, . This implies
Now consider, the ratio
Taking the limit as n tending to infinity for the absolute value of this ratio, we get as using L'Hospital's rule.
The series converges if . Hence the radius of convergence is .

c. We have So, . This implies
Now consider, the ratio
Taking the limit as n tending to infinity for the absolute value of this ratio, we get as
The series converges if . Hence the radius of convergence is .
d. We have So, . This implies
Now consider, the ratio
Taking the limit as n tending to infinity for the absolute value of this ratio, we get as for any
So the radius of convergence is zero, and there's no interval of convergence.

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