Test uniform convergence of series $\sum_{n=1}^{\infty}\frac{x}{(n+x)^2}$

Question

The question says :

Test the uniform convergence and term-by-term integration of the series $\sum_{n=1}^{\infty}\frac{x}{(n+x)^2}$

My response:

Had this been on a bounded set, series would have been uniformly convergent and hence by property, would have term-by-term integrable

But what about its uniform convergence for $\forall x \geq 0$

Answer 1

It doesn't converge uniformly. First off, it's not defined on the negative integers, but even if you exclude them, what you're trying to prove is that for every strictly positive $\epsilon$ there is an $N$ such that for every $x$ (as opposed to pointwise "for every $x$ there is an $N$") the remainder of the series beginning at $N$ is less than $\epsilon$. Therefore, to show that it doesn't converge uniformly, it suffices to show that for every $N$ there exists an $x$ such that the remainder of the series beginning at $N$ is greater than $\frac{1}{2}$. The integral test will tell you that, for positive $x$, the series beginning at $N$ is always greater than $\frac{x}{N+x}$, and you simply need to choose $x>2N$.

Answer 2

As @Math1000 commented, the exact solution for the infinite summation is $$S=\sum_{n=1}^{\infty}\frac{x}{(n+x)^2}=x\, \psi ^{(1)}(x+1)$$

If you are not requiring too much accuracy, you could have a very decent approximation building around $x=0$ the $[3,3]$ Padé approximant. This would give $$S\sim \frac{x\left(\frac{\pi^2}6 +a x+b x^2\right)} {1+c x+d x^2+e x^3}$$ The exact formulae for $(a,b,c,d,e)$ are really messy but, numerically, they are $$a=2.20248\quad b=0.824897 \quad c=2.80048\quad d=2.62052 \quad e=0.823552$$ $$\left( \begin{array}{ccc} x & \text{approximation} & \text{exact} \\ 1 & 0.644942 & 0.644934 \\ 2 & 0.789937 & 0.789868 \\ 3 & 0.851641 & 0.851469 \\ 4 & 0.885576 & 0.885292 \\ 5 & 0.907006 & 0.906615 \\ 6 & 0.921758 & 0.921271 \\ 7 & 0.932532 & 0.931959 \\ 8 & 0.940744 & 0.940096 \\ 9 & 0.947211 & 0.946497 \\ 10 & 0.952436 & 0.951663 \end{array} \right)$$

Answer 3

Note that

$$\sup_{x\in [0,\infty)} \sum_{k=n+1}^{2n} \frac{x}{(k+x)^2} \geqslant \sup_{x\in [0,\infty)} n \frac{x}{(2n+x)^2} \geqslant n \frac{n}{(3n)^2} = \frac{1}{9}$$

and the left-hand side does not converge to $0$ as $n \to \infty$.

Hence, the Cauchy criterion is violated and the series is not uniformly convergent on $[0,\infty)$.