Cfrgtkky

For which value of $delta in (0,infty)$ is $sum_{n=1}^{infty}frac{x}{n(1+n^{delta}x^{2})}$ uniformly convergent on $mathbb{R}$?

Can somebody verify my attempt, I have made quite a bit of misteps on the way but I think this should be correct:

So I have already shown that $sum_{n=1}^{infty}frac{x}{n(1+nx^{2})}$ is uniformly convergent on all of $mathbb{R}$. Because $n^{1/2}x < nx^2 +1$ (I showed this via a discriminant of a quadratic argument). First if $x >0$ it is clear that $frac{x}{n(1+nx^{2})} < frac{x}{ncdot n^{1/2}x} = frac{1}{n^{3/2}}$. If $x < 0$ then $$frac{1}{n^{1/2}} < 0 < frac{1}{nx^{2}+1} implies frac{x}{n(1+nx^{2})} < frac{x}{ncdot n^{1/2}x} = frac{1}{n^{3/2}}$$

and by the Weierstrass M-Test the series $sum_{n=1}^{infty}frac{x}{n(1+nx^{2})}$ is uniformly convergent on $mathbb{R}$.

So now I plan on doing a similar strategy.

We want to know when $n^{1/2}x < n^{delta}x^{2}+1$. As then we will have $frac{x}{n(1+n^{delta}x^{2})}< frac{1}{n^{frac{3}{2}}}$. By using the discriminant of quadratics, this happens when $n-4n^{delta}< 0$. Here we note that $1-4(1^{delta}) < 0$, for all $delta$. So let $ngeq 2$. This is when $delta > frac{-ln(4)}{ln(n)} + 1$. So since this need to hold for all $ngeq 2$, we need $delta > lim_{ntoinfty}frac{-ln(4)}{ln(n)}+1=1$. Since the biggest the RHS can be is when $ntoinfty$.

So we conclude that $sum_{n=1}^{infty}frac{x}{n(1+n^{delta}x^{2})}$ uniformly converges on $mathbb{R}$ for all $delta in [1,infty)$.

My main concern is if the series converges for any $deltain (0,1)$, as we have not compared against $1/n^p$, where $p>1$.

Notice that we have

$$
left| sum_{n=N+1}^{infty} frac{x}{n(1+n^{delta}x^2)} right|
leq int_{N}^{infty} frac{|x|}{t(1+t^{delta}x^2)} ,dt.
$$

The last integral can be computed by applying the substitution $u=x^2 t^{delta}$. (Of course, we may exclude the trivial case $x = 0$ so that the substitution makes sense.)

$$
int_{N}^{infty} frac{|x|}{t(1+t^{delta}x^2)} , dt
= frac{|x|}{delta} int_{N^{delta} x^2}^{infty} frac{du}{u(1+u)}
= frac{|x|}{delta}logleft( 1+frac{1}{N^{delta} x^2} right).
$$

It is easy to check that

$$ f(x) = begin{cases}
xlog(1+x^{-2}), & x neq 0 \
0, & x = 0
end{cases}$$

is uniformly bounded on $mathbb{R}$. Let $C > 0$ be any bound of $f$. Then

$$ forall x in mathbb{R} : quad left| sum_{n=N+1}^{infty} frac{x}{n(1+n^{delta}x^2)} right|
leq frac{|f(N^{delta/2} x)|}{delta N^{delta/2}} leq frac{C}{delta N^{delta/2}} $$

This tells that the series converges uniformly for all $delta > 0$.

By completing the square one easily obtains the following inequality $$frac{2n^{delta /2}lvert{ xrvert} }{1+n^{delta}x^{2}}leq 1 $$

Hence for all $xin mathbb{R}$, we get
$$frac{1}{n}frac{lvert x rvert}{1+ n^{delta}x^{2}}leq frac{1}{2n^{1+delta /2 }} $$
Now since $delta >0$ the uniform convergence follows from Weierstrass M-test.

Your idea works for any $delta >0$. We have $$1+n^{delta} x^{2} geq 2n^{delta /2} x$$ hence $$ frac x {n(1+n^{delta} x^{2})}leq frac 1 {2n^{1+delta/2}}$$ I have taken $x>0$ in this proof but it is obviously enogh to consider this case.