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

Question

The problem is stated as:

Examine the uniform convergence of $\sum_{n=1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)}$ on the interval $[0,\infty)$ and $[a,\infty)$ for $a>0$.

My attempt:

Case I: $[0,\infty)$

We form our partial sum $S_m(x) := \sum_{n=1}^{m} \frac{x}{(1+nx)(1+(n+1)x)}$, hence we have that:

$$ |S(x)-S_m(x)| = \sum_{n=m+1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)} $$

Picking $x_n = 1/n \in [0,1] \subset [0,\infty) \forall n \geq 1 $, we have that the series can be rewritten as:

$$ | S(x)-S_m(x)| = \sum_{n=m+1}^{\infty} \frac{1}{(2n)(2+1/n)} \geq \sum_{n=m+1}^{\infty} \frac{1}{4n+2}$$ which clearly diverges.

Therefore, we have found a counter example of uniform convergence within this interval, since if we would have uniform convergence, no matter what x we pick, the series must converge uniformly.

Case II: $[a,\infty)$

Notice that $$S(x) = \sum_{n=1}^{\infty} \frac{x}{(1+nx)(1+(n+1)x)} = \sum_{n=1}^{\infty} \left (\frac{1}{(1+2n)(1+nx)} - \frac{1}{(1+2n)(1+(n+1)x)} \right )$$

We can form our Majorant term by calculating the upper bound of the term in the series above and we have that:

$$\left |\frac{1}{(1+2n)(1+nx)} - \frac{1}{(1+2n)(1+(n+1)x)} \right | \leq \left (\frac{1}{(1+2n)(1+na)} + \frac{1}{(1+2n)(1+(n+1)a)} \right )$$

Let $M_n := \left (\frac{1}{(1+2n)(1+na)} + \frac{1}{(1+2n)(1+(n+1)a)} \right )$, and since both terms act asymptotically as $1/n^2$, it becomes evident that the series of terms given by $M_n$ does converge. By Weierstrass M - test, we have uniform convergence within this interval.

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I hope you can give me some feedback on my solution. Maybe some tips on what to improve, and what steps went wrong. I'm really trying to get good at problems with uniform convergence, so any help would be appreciated!

Thanks.

Answer 1

Note that$$\frac x{(1+nx)\bigl(1+(n+1)x\bigr)}=\frac1{1+nx}-\frac1{1+(n+1)x}$$and that therefore$$S_m(x)=\frac1{1+x}-\frac1{1+(m+1)x}.$$So, if$$S(x)=\sum_{m=1}^\infty\frac x{(1+nx)\bigl(1+(n+1)x\bigr)},$$then$$S(x)=\frac1{1+x}-\lim_{n\to\infty}\frac1{1+nx}=\begin{cases}\frac1{1+x}&\text{ if }x>0\\0&\text{ if }x=0.\end{cases}$$The convergence is not uniform on $[0,\infty)$, since$$(\forall m\in\Bbb N):S_m\left(\frac1{m+1}\right)=\frac{m+1}{m+2}-\frac12$$and$$\lim_{m\to\infty}\frac{m+1}{m+2}-\frac12=\frac12\ne0=S(0).$$Or you can say that if the convergence was uniform, then $S$ would be continuous, but that's not the case.

But the convergense is uniform on $[a,\infty)$, if $a>0$.

Answer 2

As you can see from the other answer, this problem can be simplified since the partial sums are telescoping.

However, you may be interested to see a more general approach that corrects your attempt for Case I -- particularly given your stated goal of improving your facility with problems related to uniform convergence. For uniform convergence to hold we must have

$$\tag{*}\lim_{m \to \infty} \sup_{x \in [0,\infty)}|S_m(x) - S(x)|=\lim_{m \to \infty} \sup_{x \in [0,\infty)}\left|\sum_{n=m+1}^\infty \frac{x}{(1+nx)(1+(n+1)x)}\right|=0$$

However, all terms are nonnegative for $x \in [0,\infty)$, and

$$\sup_{x \in [0,\infty)}\left|\sum_{n=m+1}^\infty \frac{x}{(1+nx)(1+(n+1)x)}\right| = \sup_{x \in [0,\infty)}\sum_{n=m+1}^\infty \frac{x}{(1+nx)(1+(n+1)x)}\\\geqslant \underbrace{\sum_{n=m+1}^{2m} \frac{x}{(1+nx)(1+(n+1)x)}}_{m \text{ terms}}\geqslant m \cdot \frac{x}{(1+2mx)(1+(2m+1)x)}$$

In particular, with $x = x_m = \frac{1}{m} \in [0,\infty)$ we have

$$\sup_{x \in [0,\infty)}\left|\sum_{n=m+1}^\infty \frac{x}{(1+nx)(1+(n+1)x)}\right|\geqslant \frac{m\frac{1}{m}}{(1+2m\frac{1}{m})(1+(2m+1)\frac{1}{m})}= \frac{1}{3(3 + \frac{1}{m})}\underset{m \to \infty}\longrightarrow \frac{1}{9}$$

Since the RHS does not converge to $0$, the necessary condition (*) does not hold and the convergence of the series is not uniform for $x \in [0,\infty)$.