I'm working through a question (it's a question with parts that builds on each other) that overall will show:

  • a series of functions does not converge uniformly on $\mathbb{R}$ by showing the sequence of its partial sums is not uniformly Cauchy on $\mathbb{R}$

The series of functions is: $$\sum_{n=1}^{\infty}{\dfrac{1}{\sqrt{n}}\sin\left(\frac{x}n\right)} \quad $$

Firstly, I'm asked to show the negation for uniformly Cauchy:

  • For some $\epsilon > 0$, for all $n$ $\epsilon$ $\mathbb{N}$, there exists an $m, n \geq N$ and there exists an $x$ $\epsilon$ I such that $|f_m{(x)} - f_n{(x)}| \geq \epsilon$

Secondly: If $k, N$ $\epsilon$ $\mathbb{N}$ and $N + 1 \leq k \leq 4N + 3$, show that :

  • $1 \geq \sin\left(\frac{N+1}k\right) \geq \frac{1}5$

I have done this on paper, but would take while to write out, so will omit my answer here.

Third (where I'm beginning to get stuck): Show that for all $N$ $\in$ $\mathbb{N}$ $$\sum_{k=N+1}^{k=4N+3}{\dfrac{1}{\sqrt{k}}} \geq 2\sqrt{2}$$

My idea for part 3 is to compare it to an integral, specifically the integral of $\frac1{\sqrt{x}}$, but I'm not sure how to show this relation for all $N$ (what the limits of integration should be) - any tips?


  • Show that when $x = N + 1, m = 4N + 3$, and $n = N$, $$|S_m{(x)} - S_n{(x)}| \geq \frac{2\sqrt{2}}5$$

    And then ultimately show the original sequence doesn't uniformly converge on $\mathbb{R}$

Part C I think I'm on the right track and some tips should suffice, but for part d I'm relatively lost. Any help's greatly appreciated - thank you


Part 2

Let $N + 1 \le k \le 4N + 3$.

Then $\frac{N+1}{k}$ satisfies the inequalities $\frac{N+1}{4N+3}\le\frac{N+1}{k}\le1$.

Now, for $x$ in the interval $\left(\frac{N+1}{4N+3},1\right)$, $\sin x$ is monotonically increasing and we have

$$1\ge \sin(1)\ge \sin\left(\frac{N+1}{k}\right)\ge\sin\left(\frac{N+1}{4N+3}\right)$$

and thus

$$\begin{align} \sin\left(\frac{N+1}{k}\right)&\ge\sin\left(\frac{N+1}{4N+3}\right)\\\\ &\ge\sin(1/4)\\\\ &\ge \frac14-\frac{1}{3!}(\frac14)^3\\\\ &=\frac14\left(1-\frac{1}{96}\right)\\\\ &>\frac15 \end{align}$$

which was to be shown.

Part 3

For this part, we will use the result from the following integral

$$\begin{align} \int_{N+1}^{4N+4}\frac{dx}{\sqrt{x}}&=2\left(\sqrt{4N+4}-\sqrt{N+1}\right)\\ &=2\sqrt{N+1}(2-1)\\ &\ge 2\sqrt{2} \end{align}$$

for $N\ge1$.

Now, we note that this integral can be represented by the following summation:

$$\begin{align} \int_{N+1}^{4N+4}\frac{dx}{\sqrt{x}}&=\sum_{k=N+1}^{4N+3}\int_{k}^{k+1} \frac{dx}{\sqrt{x}}\\ &\le\sum_{k=N+1}^{4N+3} \frac{1}{\sqrt{k}} &\ge 2\sqrt{2} \end{align}$$

and we have the desired inequality!

Part 4

From the Part 2, we showed that if $N + 1 \le k \le 4N + 3$, then

$$1 \ge \sin\left(\frac{N+1}{k}\right) \ge \frac15$$

From the Part 3, we showed that for all $N$

$$\sum_{k=N+1}^{4N+3}{\frac{1}{\sqrt{k}}} \ge 2\sqrt{2}$$

Then, putting these together we have

$$\begin{align} |S_m(x)-S_n(x)|&=\left|\sum_{k=N+1}^{4N+3} \frac{\sin((N+1)/k)}{\sqrt{k}}\right|\\ &\ge\left|\sum_{k=N+1}^{4N+3} \frac{\frac15}{\sqrt{k}}\right|\\\\ &\ge\frac15\left|\sum_{k=N+1}^{4N+3} \frac{1}{\sqrt{k}}\right|\\\\ &\ge\frac{2\sqrt{2}}{5} \end{align}$$


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