Prove uniform convergence point by point?

Question

I want to prove series $\sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{x+k}\right)$ converges uniformly for $x\geq0$. By Weirstrass test, i want to find sequence $M_n$ that satisfies

1. $|f_n(x)| \leq M_n$ for every $n\geq0$, $x\geq0$,
2. $\sum_{n=1}^\infty M_n$ converges.

So far i got for $x\geq0$

$|\frac{1}{k}-\frac{1}{x+k}|=|\frac{x+k-k}{k(x+k)}|=|\frac{x}{k(x+k)}|=\frac{x}{k(x+k)}=\frac{x}{kx+k^2} \leq \frac{x}{k^2}$,

but i struggle to find sequence independent of $x$. Is it legit to define

$M_k(x):=\frac{x}{k^2}$

and apply the m- test for every point $x\geq0$ since

for $x\geq0$: $|f_n(x)| \leq M_n(x)$, $n\geq0$ and $\sum_{n=1}^\infty M_n(x)$ converges?

Is there any "easier" sequence to bound my series?

Answer 1

Your series is uniformly convergent on any interval $\,[0,a]\,,\,$ where $\,a>0\,,\,$ indeed, for all $\,a>0\,,\,$ it results that

$\displaystyle\sum_{k=1}^{\infty}\dfrac a{k^2}\;$ is a convergent series and

$\left|\dfrac{1}{k}\!-\!\dfrac{1}{x+k}\right|=\left|\dfrac{x+k-k}{k(x+k)}\right|=\left|\dfrac{x}{k(x+k)}\right|=\dfrac{x}{k(x+k)}\leqslant\dfrac{a}{k^2}$

for any $\,k\in\Bbb N\,$ and for any $\,x\in[0,a]\,.$

Now, I will prove by contradiction that your series is not uniformly convergent on the interval $\,[0,+\infty)\,.$

If it were uniformly convergent on $\,[0,+\infty)\,,\,$ by using Cauchy’s criterion, it would follow that

for $\;\varepsilon=\dfrac12>0\,,\,$ there would exist $\,\nu\in\Bbb N\,$ such that

$\displaystyle\left|\sum_{k=n}^{m-1}\left(\dfrac1k-\dfrac1{x+k}\right)\right|<\dfrac12\;$ $\;\forall \,m\!>\!n\!>\!\nu\;$ and $\;\forall x\in[0,+\infty)\,.$

In particular, for $\,n\!=\!\nu\!+\!1\,$ and $\,m\!=\!2\nu\!+\!2\,,\,$ we would get that

$\displaystyle\left|\sum_{k=\nu+1}^{2\nu+1}\left(\dfrac1k-\dfrac1{x+k}\right)\right|=\!\!\sum_{k=\nu+1}^{2\nu+1}\dfrac{x}{k(x+k)}<\dfrac12\;$ $\;\forall x\!\in\![0,+\infty)\,,$

consequently,

$\displaystyle\lim_{x\to+\infty}\left[\sum_{k=\nu+1}^{2\nu+1}\dfrac{x}{k(x+k)}\right]=\sum_{k=\nu+1}^{2\nu+1}\dfrac1k\leqslant\dfrac12\;\;,$

but it leads to a contradiction, indeed

$\displaystyle\sum_{k=\nu+1}^{2\nu+1}\dfrac1k>\sum_{k=\nu+1}^{2\nu}\dfrac1k\geqslant\nu\!\cdot\!\dfrac1{2\nu}=\dfrac12\;.$

Hence, your series is not uniformly convergent on the interval $\,[0,+\infty)\,.$

Answer 2

The series fails to converge uniformly for all $x \geqslant 0$ since

$$\sup_{x \geqslant 0}\left|\sum_{k=n+1}^\infty \frac{x}{k(x+k)}\right| = \sup_{x \geqslant 0}\sum_{k=n+1}^\infty \frac{x}{k(x+k)}\geqslant \sup_{x \geqslant 0}\sum_{k=n+1}^{2n} \frac{x}{k(x+k)}\\\geqslant \sum_{k=n+1}^{2n}\frac{n}{k(n+k)}\geqslant n \cdot \frac{n}{2n(n+2n)} = \frac{1}{6}\underset{n \to \infty}\longrightarrow \frac{1}{6} \neq 0$$

Answer 3

The convergence of this series of positive increasing functions is uniform on $[0,A]$ for every positive $A$ as you already noticed, but not on $[0,+\infty).$ Even worse: for every $n\in\Bbb N,$ $$\sup_{x\ge0}\sum_{k\ge n}f_k(x)=\sum_{k\ge n}\lim_{+\infty}f_k=\sum_{k\ge n}\frac1k=+\infty.$$