# uniform convergence and proof

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 sinnx ) is uniformly convergent, then lim(nk_n )=0.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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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