Using Dirichlet's test for uniform convergence

[cchatham]

Homework Statement

Show that $$\sum$$(-1)^(n+1) / (n + x^2) converges uniformly but not absolutely on R.

Homework Equations

Using Dirichlet's Test for uniform convergence. (fn) and (gn) are sequences of functions on D satisfying:
$$\sum$$ fn has uniformly bounded partial sums
gn -> 0 uniformly on D
g(n+1)(x) $$\leq$$ gn(x) for all x in D and all n in N

The Attempt at a Solution

I got parts a) and c) down since obviously (-1)^(n+1) is uniformly bounded by 1 and 1 / (n+1+x^2) < 1 / (n + x^2) for all x in R and all n in N. I'm having trouble showing that 1 / (n+x^2) is uniformly convergent on R. I tried using Weierstrass' M-test but can't seem to make that work.

 

[mighty2000]

Thank you for your post.

To show that \sum(-1)^(n+1) / (n + x^2) converges uniformly on R, we can use Dirichlet's Test for uniform convergence.

First, we can see that (-1)^(n+1) is uniformly bounded by 1 for all n in N and all x in R.

Next, we need to show that the sequence of functions (1 / (n + x^2)) converges to 0 uniformly on R. To do this, we can use the Weierstrass M-test.

Let M_n = 1 / (n + x^2). Then, for all x in R and all n in N, we have M_n < 1 / n. Since \sum 1 / n converges, by the Weierstrass M-test, \sum M_n also converges uniformly on R.

Finally, we need to show that g(n+1)(x) \leq gn(x) for all x in R and all n in N.

Let g_n(x) = 1 / (n + x^2). Then, g(n+1)(x) = 1 / (n+1 + x^2) and gn(x) = 1 / (n + x^2). Since n+1 > n, we have g(n+1)(x) \leq gn(x) for all x in R and all n in N.

Therefore, by Dirichlet's Test for uniform convergence, we can conclude that \sum(-1)^(n+1) / (n + x^2) converges uniformly on R.

To show that the series does not converge absolutely on R, we can use the Cauchy Condensation Test.

Let a_n = 1 / (n + x^2). Then, a_n+1 = 1 / (n+1 + x^2).

By the Cauchy Condensation Test, if \sum 2^n a_2^n converges, then \sum a_n also converges.

But \sum 2^n a_2^n = \sum 2^n / (2^n + x^2).

Since 2^n + x^2 > 2^n for all x in R and all n in N, we have