Verified answer

For each x, we have an alternating series. Since 1/(n + x^2) → 0 for every x, then the series Σ(n=1 to ∞) (-1)^n) /(n+x^2) converges for every x in [0, ∞), which is the same as to say that this series of functions converges on [0, ∞).

To show the convergence is uniform, let S_n be the n_th partial sum of the given series and let S be its limit. Since for each x we have a convergent alternating series, then, for each n and each x we have

|S_n(x) - S(x)| < 1/(n + 1 + x^2) ≤ 1/(n + 1)

Therefore, given ε > 0, if we choose n > 1/ε - 1 we have |S_n(x) - S(x)| < ε for every x. Then, the convergence is uniform on [0, ∞).

On the other hand, if we take Σ(n=1 to ∞) |(-1)^n) /(n+x^2)| = Σ(n=1 to ∞) 1/(n+x^2), then, for every x, lim (1/(n + x^2))/(1/n) = lim n/(n + x^2) = 1. Since the harmonic series Σ(n=1 to ∞) 1/n diverges to ∞ then so does Σ(n=1 to ∞) 1/(n+x^2). Therefore, the given series is not absolutely convergent.

The series is uniformly convergent if the rate of convergence is uniform, meaning: for any d > 0, there exists N>0 such that the difference between the kth partial sum and the limit is less than d for all x, for all k > N. In other words, the rate of convergence (represented by what N is chosen for a given d) does not depend on the value of x. For this series the slowest convergence occurs when x is 0, so if you can figure out what N works for a given d>0 and for x=0 then I promise that will work for all x>0. For x=0 this is the alternating harmonic series. It doesn't converge absolutely for the same reason (basically) that the harmonic series doesn't converge. This can be done by series comparison for any given x - the tail of the harmonic series doesn't converge, and using the floor function on x there is a term by term comparison with the tail of the harmonic series.

If x >= a > 0, then a million / (a million + n^2x) <= a million / (a million + n^2a) < a million/(n^2a). The sequence whose words are a million / (n^2a) converges by utilising the p-sequence try. considering that this convergence does not remember on x, that's uniformly convergent. - - - - to make certain it is not uniformly convergent on (0,inf), word that given n, we could set x = a million/n^2 to get a million / (a million + n^2x) = a million/2. for this reason the words do no longer converge to 0 uniformly on (0,inf)