Let $f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n} \arctan\left(\frac{x}{\sqrt n}\right)$. Show that $f(x)$ converges uniformly.

First, it is easy to see that the series converges for every $x$ by Leibniz test.

Now, I'm not so sure how to prove uniform converges. I thought about the fact that $\arctan\left(\frac{x}{\sqrt n}\right)$ is bounded by $\frac{\pi}{2}$, problem is, it's not a supremum but an upper-bound.

I've tried to look for other tests like Weierstrass M-test but it didn't fit here.

I think we should use Cauchy criteria. Let's assume by contradiction it is diverges there is an $\varepsilon > 0$ such that for every $N$ there are $m,n > N$ such that:

$$\left| \sum_{k=m}^n \frac{(-1)^k}{\sqrt k} \arctan(\frac{x}{\sqrt k})\right| \ge \varepsilon$$

Now, for every $x$:

$$\left| \sum_{k=m}^n \frac{(-1)^k}{\sqrt k} \arctan(\frac{x}{\sqrt k})\right| \le \left| \sum_{k=m}^n \frac{(-1)^k}{\sqrt k} \frac{\pi}{2} \right| = \frac{\pi}{2} \left|\sum_{k=m}^n \frac{(-1)^k}{\sqrt k}\right|$$

Since the later series converges by Leibnitz test, it is a Cauchy series and therefore there is an $N$ such that for every $m,n$:

$$\left|\sum_{k=m}^n \frac{(-1)^k}{\sqrt k}\right| < \frac{2\varepsilon}{\pi}$$

And so, we're done.

Could you verify my proof please?


  • $\begingroup$ This may help. $\endgroup$ – David Mitra Feb 28 '15 at 15:48
  • $\begingroup$ @DavidMitra, I am actually aware of this but we haven't learned this, so I am looking for a good alternative. $\endgroup$ – AlonAlon Feb 28 '15 at 15:49
  • $\begingroup$ Or maybe the best option is to imitate this proof to my specific case... $\endgroup$ – AlonAlon Feb 28 '15 at 15:51

The inequality

$$ \left | \sum_{k=m}^{n} \frac{(-1)^k}{\sqrt{k}}\arctan{\frac{x}{\sqrt{k}}} \right | \le \left | \sum_{k=m}^{n} \frac{(-1)^k}{\sqrt{k}}\frac{\pi}{2} \right | $$

Does not need to be true generally since the series is alternating. In this case it probably is true but you should also show that as well.
We know definitely that

$$ \left | \sum_{k=m}^{n} \frac{(-1)^k}{\sqrt{k}}\arctan{\frac{x}{\sqrt{k}}} \right | \le \sum_{k=m}^{n} \left | \frac{(-1)^k}{\sqrt{k}}\frac{\pi}{2} \right | \le \frac{\pi}{2} \sum_{k=m}^{n}\frac{1}{\sqrt{k}} $$

But that is not sufficient as $\sum_{k=1}^{+\infty} \frac{1}{\sqrt{k}}$ is not converging.

However there is an easier way.

To prove your uniform convergence i will use Abel's uniform convergence test And since the domain is unknown i will assume that $x \in \mathbb{R}$

i) Series $\sum_{k=1}^{+\infty} \frac{(-1)^k}{\sqrt{k}}$ is converging by Leibniz rule

ii) $\arctan{\frac{x}{\sqrt{k}}}$ is uniformly bounded sequence on $(-\infty,+\infty)$

$$ (\exists M>0)(\forall x \in (-\infty,+\infty))(\forall k \in \mathbb{N}) \left | \arctan{\frac{x}{\sqrt{k}}} \right | < M $$

In our case $M=\frac{\pi}{2}$

iii) $\arctan{\frac{x}{\sqrt{k}}}$ is monotonous on

$x \in [0,+\infty)$ Decreasing by n

$x \in (-\infty,0]$ Increasing by n

By i) , ii) and iii) We deduct that begining functional series in uniformly convergent on $(-\infty,0] \cup[0,+\infty)$


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