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I need help with the following problem.

Consider the serie of function

[tex]\sum_{n=1}^{\infty}\frac{1}{1+n^2x}[/tex]

The serie is undefined for [itex]x \in \{0\}\cup \{-1/n^2, \ n\in \mathbb{N}\}[/itex]. I want to find wheter it converges pointwise in (-1, 0) or not and if it does, does it converge uniformly?

The way I would start this problem is by saying: For a given number [itex]m \in \mathbb{N}[/itex], consider

[tex]x_0 \in \left(\frac{-1}{m^2} \ ,\frac{-1}{(m+1)^2}\right)[/tex]

Consider

[tex]f_n(x) = \frac{1}{1+n^2x}[/tex]

Then

[tex]|f_n(x_0)| = \frac{1}{|1+n^2x_0|} = \frac{1}{|1-n^2|x_0||}= \left\{ \begin{array}{rcl}

\frac{1}{1-n^2|x_0|} & \mbox{for}

& n<\sqrt{\frac{1}{|x_0|} \\

\frac{1}{n^2|x_0|-1} & \mbox{for}

& n>\sqrt{\frac{1}{|x_0|}

\end{array}\right [/tex]

and

[tex]\sum_{n=1}^{\infty}|f_n(x_0)| = \sum_{n=1}^{\left[\sqrt{1/|x_0|}\right]}\frac{1}{1-n^2|x_0|} + \sum_{n=\left[\sqrt{1/|x_0|}\right]+1}^{\infty}\frac{1}{n^2|x_0|-1}[/tex]

I'm guessing this serie converges, but I'm having trouble finding a convergent serie to bound it with. The other convergence tests have failed and the use of the integral convergence criterion is forbiden. I know that if there is a serie to bound it with, it would be of the form

[tex]\sum_{n=1}^{\infty}a_n = \sum_{n=1}^{\left[\sqrt{1/|x_0|}\right]}\frac{1}{1-n^2|x_0|} + \sum_{n=\left[\sqrt{1/|x_0|}\right]+1}^{\infty}b_n[/tex]

with

[tex]\frac{1}{n^2|x_0|-1} \leq b_n[/tex]

for n > N.

Edit:

And if there exists such an N that also satisfies

[tex]N\leq \left[\sqrt{1/|x_0|}\right][/tex]

then according to Weirstrass M-test, the convergence is uniform.

Consider the serie of function

[tex]\sum_{n=1}^{\infty}\frac{1}{1+n^2x}[/tex]

The serie is undefined for [itex]x \in \{0\}\cup \{-1/n^2, \ n\in \mathbb{N}\}[/itex]. I want to find wheter it converges pointwise in (-1, 0) or not and if it does, does it converge uniformly?

The way I would start this problem is by saying: For a given number [itex]m \in \mathbb{N}[/itex], consider

[tex]x_0 \in \left(\frac{-1}{m^2} \ ,\frac{-1}{(m+1)^2}\right)[/tex]

Consider

[tex]f_n(x) = \frac{1}{1+n^2x}[/tex]

Then

[tex]|f_n(x_0)| = \frac{1}{|1+n^2x_0|} = \frac{1}{|1-n^2|x_0||}= \left\{ \begin{array}{rcl}

\frac{1}{1-n^2|x_0|} & \mbox{for}

& n<\sqrt{\frac{1}{|x_0|} \\

\frac{1}{n^2|x_0|-1} & \mbox{for}

& n>\sqrt{\frac{1}{|x_0|}

\end{array}\right [/tex]

and

[tex]\sum_{n=1}^{\infty}|f_n(x_0)| = \sum_{n=1}^{\left[\sqrt{1/|x_0|}\right]}\frac{1}{1-n^2|x_0|} + \sum_{n=\left[\sqrt{1/|x_0|}\right]+1}^{\infty}\frac{1}{n^2|x_0|-1}[/tex]

I'm guessing this serie converges, but I'm having trouble finding a convergent serie to bound it with. The other convergence tests have failed and the use of the integral convergence criterion is forbiden. I know that if there is a serie to bound it with, it would be of the form

[tex]\sum_{n=1}^{\infty}a_n = \sum_{n=1}^{\left[\sqrt{1/|x_0|}\right]}\frac{1}{1-n^2|x_0|} + \sum_{n=\left[\sqrt{1/|x_0|}\right]+1}^{\infty}b_n[/tex]

with

[tex]\frac{1}{n^2|x_0|-1} \leq b_n[/tex]

for n > N.

Edit:

And if there exists such an N that also satisfies

[tex]N\leq \left[\sqrt{1/|x_0|}\right][/tex]

then according to Weirstrass M-test, the convergence is uniform.

Last edited: