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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.

    © BrainMass Inc. brainmass.com October 4, 2022, 11:30 am ad1c9bdddf
    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.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 4, 2022, 11:30 am ad1c9bdddf>
    https://brainmass.com/math/real-analysis/uniform-convergence-proof-297769

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