Examine uniform convergence of the series $$\sum_{n=1}^\infty \frac{x}{(1+(n-1)x)(1+nx)}$$

on the intervals $[a,b]$ where $(0 < a < b)$ and $[0,b]$ where $(b>0)$

My attempt:

(i) On the interval $[a,b]$

Approach (1)

Notice that $$\frac{x}{(1+(n-1)x)(1+nx)} = \frac{1}{1+(n-1)x} - \frac{1}{1+nx}$$

Hence: $$\sum_{k=1}^n \frac{x}{(1+(n-1)x)(1+nx)} = 1 - \frac{1}{1+nx}$$

so the series converges pointwise to the constant $1$-function.

The convergence is uniform:

$$\sup_{x \in [a,b]}\left|1-\frac{1}{1+nx}-1\right| \leq \frac{1}{na} \to 0$$

Approach (2)

$$\sup_{x \in [a,b]} \left|\frac{x}{(1+(n-1)x)(1+nx)}\right| \leq \frac{b}{(1+(n-1)a)(1+na)} = \frac{b}{n²}\frac{1}{(1/n + a((n-1)/n))(1/n +a)}$$

and the last expression is a converging sequence (to $a^{-2}$), so it is bounded above by a real number $M$

and hence:$$\sup_{x \in [a,b]} \left|\frac{x}{(1+(n-1)x)(1+nx)}\right| \leq \frac{b}{(1+(n-1)a)(1+na)} \leq \frac{Mb}{n^2}$$

and we can deduce uniform convergence by the Weierstrass' M-test.

(ii) On the interval $[0,b]$

Approach (1)

Again, using the partial sums, we can see that the series of functions converges pointwise to $f(x) = \begin{cases}1 \quad x \neq 0 \\0 \quad x =0\end{cases}$. This is a function that is not continuous, but the partial sums are. Hence, the convergence can't be uniform.

Approach (2)

We can also show this directly. Let $n \in \mathbb{N}$. Let $m \geq n$ such that $\frac{1}{m}< b$. Let $x = \frac{1}{m}$. Then:

$$\left|1-\frac{1}{1+mx}-1\right| = 1/2$$

Hence, the series does not converge uniformly.

Approach (3)

Now, I would like to use Cauchy's convergence criterium, as I consider using partial sums a bit cheating, in the sense that it is often impossible to find them.

However, I'm struggling with this one.

Can someone give a proof that the series does not converge uniformly with the cauchy criterium?

I.e., prove that:

$$\exists \epsilon > 0: \forall n : \exists p,q > n: \exists x \in [0,b] \left|\sum_k^p - \sum_k^q\right| \geq \epsilon$$

So my questions:

How do I complete the proof with the Cauchy criterium? Is what I wrote down correct?

  • 4
    $\begingroup$ I just want to say that this is one of the best-asked homework questions I've seen at math.SE in a long time. $\endgroup$ – Steven Stadnicki Jan 2 '18 at 23:54
  • $\begingroup$ Thank you. That's a nice compliment! $\endgroup$ – user370967 Jan 2 '18 at 23:55
  • $\begingroup$ I didn't try if it helps, but you can try bounding $|\sum_k^p-\sum_k^q|$ from below in a nontrivial way, so not something like $0$ as a bound.. $\endgroup$ – Shashi Jan 3 '18 at 11:39
  • $\begingroup$ I tried that, but didn't find anything useful. $\endgroup$ – user370967 Jan 3 '18 at 11:59
  • $\begingroup$ If you can write this out as an answer I will gladly upvote and accept. $\endgroup$ – user370967 Jan 3 '18 at 13:17

choose $p=N+1$, $q=2N$ and $x=\frac1N$ then $$\sum_{n=p}^{q}\frac{x}{(1+(n-1)x)(1+nx)}=\sum_{n=N+1}^{2N}\left(\frac1{1+(n-1)x}-\frac1{1+nx}\right)=\frac1{1+Nx}-\frac1{1+2Nx}=\frac{Nx}{(1+Nx)(1+2Nx)}=\frac16$$ so if we let $\frac16>\varepsilon > 0$ and for any $N>n$(when $n$ sufficiently large such that $\frac1n<b$ so that $x$ still in its domain), the previous equality always true.


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