Show that $\sum\limits_{k=1}^{+\infty}[\frac{1}{k}-\frac{1}{k+t}]$ is uniformly-convergent

Question

Studying notes about Gamma function, I came across the following problem that I was unable to solve.

Let's be $\phi = \frac{d}{dx}(\log(F(x))),$ where $F(x) = \int_{0}^{+\infty} t^{x}e^{-t}dt$,

Forgetting for a second that we defined $\phi$ in the way above,

We were able with a lot of calculation to find that $\displaystyle\phi = \sum_{k=1}^{+\infty}\left[\frac{1}{k}-\frac{1}{k+t}\right]$ (which is convergent by comparison)

But we also found a function $\Phi$, $\displaystyle\Phi = \sum_{k=1}^{+\infty}\left[\frac{x}{k}-\log\left(1+\frac{x}{k}\right)\right]$ such that $\Phi'=\phi$.

The problem is that I'd like to solve is to show without any bigger theorem, just with calculations, that $$\sum_{k=1}^{+\infty}\left[\frac{1}{k}-\frac{1}{k+t}\right]$$converges uniformly under the uniform norm, on every $[a,b] \subseteq ]-1,+\infty[$, or equivalently that $\left\lVert \phi - \phi_{m}\right\rVert_{\infty,[a,b]} \underset{m\to \infty}{\longmapsto} 0$, where $\phi_{m}$ is the sum up to $m$.

In this way I think I could justify the following:

\begin{align} \int_{0}^{x}\phi(t)dt &= \int_{0}^{x}\sum_{k=1}^{+\infty}\left[\frac{1}{k}-\frac{1}{k+t}\right]dt \\&= \sum_{k=1}^{+\infty}\int_{0}^{x} \left[\frac{1}{k}-\frac{1}{k+t}\right]dt \\&= \sum_{k=1}^{+\infty}\left[\frac{x}{k}-\log\left(1+\frac{x}{k}\right)\right] \\&= \Phi(x) \end{align}

In other words that $$\int_{0}^{x}\phi(t)dt = \int_{0}^{x}\Phi'(t)dt = \Phi(x)-\Phi(0)=\Phi(x).$$

Is there a relatively simple way to do it?

Answer 1

Your $k$th summand equals $\dfrac{t}{k(k+t)}.$ We can thus write the series as

$$\sum_{k=1}^{\infty}\frac{t}{k(k+t)}.$$

This series converges uniformly on $[a,b]\subset (-1,\infty)$ iff the sum from $k=2$ to $\infty$ converges uniformly on $[a,b].$ For this range of $k,$ we can use Weierstrass M: Fix $-1<a<b.$ Then for $k\ge 2$ and $t\in [a,b]$ we have

$$\left |\frac{t}{k(k+t)}\right| \le \frac{|t|}{k(k-1)} \le \frac{|a|+|b|}{(k-1)^2}.$$

Since

$$\sum_{k=2}^{\infty}\frac{|a|+|b|}{(k-1)^2} = \frac{(|a|+|b|)\pi^2}{6} <\infty,$$

we have uniform convergence on $[a,b]$ by Weierstrass M.

Answer 2

It is not uniformly convergent on $\mathbb R$. Suppose it is. Then there exists $N$ such that $N_2 >N_1 >N$ implies $\sum\limits_{k=N_1}^{N_2} \frac t {k(k+t)} <1$ for all $t$. Put $t=N_2$ and observe that $\frac {N_2} {k(k+N_2)} \geq \frac 1 {2k}$ whenever $N_1 \leq k \leq N_2$. This gives $\sum\limits_{k=N_1}^{N_2} \frac 1 {2k} <1$ whenever $N_2 >N_1 >N$. But this means that the harmonic d=series is convergent!