Uniform convergence of the series on unbounded domain

Question

1 $$\sum_{n=1}^\infty (-1)^n \frac{x^2 + n}{n^2} $$ Is the series converges uniformly $\mathbb R$

I have tired by this result

• if $\{f_n(x)\}$ is a sequence of a function defined on a domain $D$ such that

  1. $f_n(x) \geq 0$ for all $x \in D$ and for all $n \in \mathbb N$

  2. $f_{n+1}(x) \leq f_n(x)$ for all $x \in D$

  3. $\sup_{x\in D} \{f_n(x)\} \to 0$ as $n \to \infty$. Then

$\sum_{n=1}^\infty (-1)^{(n+1)}f_n(x)$ converges uniformly on $D$

first two condition this series is satisfied but third is not satisfied, so this is not work

2. $$ \sum_{n=1}^\infty \frac{x \sin \sqrt{\frac{x}n}}{x +n} $$ Is the series converges uniformly on $[1, +\infty)$

I have tried Abel test and Dirichlet test,but not getting any solution.

3. Study the uniform convergence of the series on $\mathbb R$ $$ \sum_{n=1}^\infty \frac{x\sin(n^2x)}{n^2} $$

Suppose this series converges uniformlybto $f$, so for a given $\epsilon > 0$, there exist $N \in \mathbb N$ such that $\left|\sum_{n=1}^k f_n(x) - f(x)\right| < \epsilon$ for all $k \geq N$

Please tell me how to proceed further. Any help would be appreciated , Thank you

Answer 1

1. In general, if $\sum f_n(x)$ converges uniformly on $\mathbb R,$ then

$$ \sup_{\mathbb R}|f_n| \to 0 \text { as } n\to \infty.$$

This fails in your problem, because in this case $\sup |f_n| \ge |f_n(n^2)| = 1+1/n \not \to 0.$

2. We can use the same idea as in 1. In this problem, $f_n(n) = (1/2)\sin 1 \not \to 0,$ so the series doesn't converge uniformly on $[1,\infty).$

3. The series converges uniformly on any $[-a,a].$ Proof: We use Weierstrass M:

$$\sup_{[-a,a]}|f_n| \le \frac{a\cdot 1}{n^2}.$$

Since $\sum a/n^2 < \infty$ we have $\sum f_n$ converging uniformly on $[-a,a].$

However, the series does not converge uniformly on $\mathbb R.$ We can again use the idea in 1. Let $x_n = \pi/2n^2 + 2\pi n^2.$ Then verify

$$\sup_{\mathbb R} |f_n| \ge |f_n(x_n)|\ge 1 \not \to 0.$$

Answer 2

Regarding the first series $$\sum_{n=1}^\infty (-1)^n \frac{x^2 + n}{n^2} $$ the remainder $$R_{2n}(x) = \sum_{k=2n}^\infty (-1)^n \frac{x^2 + k}{k^2}$$ can be evaluated grouping one even term with one odd term. You get $$R_{2n}(x) = x^2\sum_{k=2n}^\infty \frac{2k+1}{k^2(k+1)^2} + \sum_{k=2n}^\infty \frac{1}{k(k+1)}$$ The second term of the RHS is converging to $0$ as the series $\sum \frac {1}{k(k+1)}$ is convergent.

Regarding the first one, we have $\frac{2k+1}{k^2(k+1)^2} \sim \frac{2}{k^3}$. Hence $$\sum_{k=2n}^\infty \frac{2k+1}{k^2(k+1)^2} \sim \frac{A}{n^2}$$ with $A > 0$. Therefore $R_{2n}(x)$ doesn't converge uniformly to zero (consider $x=n$). And the series is not uniformly convergent.

Answer 3

For the second problem, we can use the inequality

$$\sin\left(\sqrt{\frac{x}{n}}\right)\ge \sqrt{\frac{x}{x+n}}$$

for $0\le \sqrt{x/n}\le \pi/2$.

Then, we have

$$\begin{align} \sum_{n=N}^\infty \frac{x\,\sin\left(\sqrt{\frac{x}{n}}\right)}{x+n}&\ge \sum_{n=N}^\infty \left(\frac{x}{x+n}\right)^{3/2}\\\\ &\ge \int_N^\infty \left(\frac{x}{x+y}\right)^{3/2}\,dy\\\\ &=\frac{2x^{3/2}}{\sqrt{x+N}}\\\\ &>1 \end{align}$$

whenever $x=N$ for $N\ge1$.

Therefore, there exists a number $\epsilon>0$ (here $\epsilon=1$ is suitable) so that for all $N'$ there exists an $x\in[1,\infty)$ (here $x=N$), and there exists a number $N>N'$ (here, take any $N>N'\ge 1$) such that $\left|\sum_{n=N}^\infty \frac{x\,\sin\left(\sqrt{\frac{x}{n}}\right)}{x+n}\right|\ge \epsilon$.

This is the statement of negation of uniform convergence and therefore the series fails to converge uniformly.