How to show that $\frac{d}{dx}\sum_{n=1}^\infty\left(\frac xn-\ln\left(1+\frac xn\right)\right)=H_x$

Question

So I was trying to find a series expansion of $\Gamma(x+1)$ (which is the analytic continuation of $x!$ when I bumped into this sum $$\lambda(x)=\sum_{n=1}^\infty\left(\frac xn-\ln\left(1+\frac xn\right)\right)$$According to Desmos, the derivative of this function is the harmonic series $H_x$. The analytic continuation of it is $$\int_0^1\frac{t^x-1}{t-1}dt$$So I want to show that $$\lambda'(x)=H_x$$So I start like this: $$\frac{d}{dx}\sum_{n=1}^\infty\left(\frac xn-\ln\left(1+\frac xn\right)\right)=\lambda'(x)$$Then we switch the derivative and sum symbol: $$\sum_{n=1}^\infty\frac{d}{dx}\left(\frac xn-\ln\left(1+\frac xn\right)\right)$$And then everything is straightforward until we arrive at $$\sum_{n=1}^\infty\left(\frac1n-\frac1{n+x}\right)$$Which could be proven that it telescopes. So did I do everything correctly so far? I am a bit suspicious about switching the summation symbol and derivative symbol at the beginning. But numerically this is correct.

And if you are wondering what the series expansion of $\Gamma(x+1)$ looks like, then here it is:$$\Gamma(x+1)=\sum_{n\ge0}\frac{\left(-\gamma x+\sum_{k=1}^\infty\frac xk-\ln\left(1+\frac xk\right)\right)^n}{n!}$$ Where $\gamma$ is Euler's constant. An interesting note is that $$\lambda(1)=\gamma$$

Answer 1

$$\frac{d}{dx}\sum_{n=1}^\infty\left(\frac xn-\ln\left(1+\frac xn\right)\right)=\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+x}\right)$$

$$=\lim_{N\to \infty}\sum_{n=1}^N\left(\frac{1}{n}-\frac{1}{n+x}\right)$$

$$=\lim_{N\to \infty}\left[\sum_{n=1}^N\frac{1}{n}-\sum_{n=1}^N\frac{1}{n+x}\right]$$

$$=\lim_{N\to \infty}\left[\sum_{n=1}^N\frac{1}{n}-\left(\underbrace{\sum_{n=1}^{N-x}\frac{1}{n+x}}_{n\to n-x}+\underbrace{\sum_{n=N-x+1}^{N}\frac{1}{n+x}}_{n\to N-n}\right)\right]$$

$$=\lim_{N\to \infty}\left[\sum_{n=1}^N\frac{1}{n}-\sum_{n=x+1}^{N}\frac{1}{n}-\sum_{n=0}^{x-1}\frac{1}{N+x-n}\right]$$

$$=\lim_{N\to \infty}\left(\sum_{n=1}^N\frac{1}{n}-\sum_{n=x+1}^{N}\frac{1}{n}\right)-\lim_{N\to \infty}\sum_{n=0}^{x-1}\frac{1}{N+x-n}$$

$$=\lim_{N\to \infty}\left(\sum_{n=1}^x\frac{1}{n}\right)-\lim_{N\to \infty}\sum_{n=0}^{x-1}\frac{1}{N+x-n}$$

$$=\sum_{n=1}^x\frac{1}{n}-0$$

$$=H_x$$

Answer 2

Hoping that it not too advanced for the time being, consider $$A_p(x)=\sum_{n=1}^\infty\left(\frac xn-\log\left(1+\frac xn\right)\right)$$ $$A_p(x)=x H_p+\log (\Gamma (p+1))-\left(\zeta ^{(1,0)}(0,p+x+1)-\zeta ^{(1,0)}(0,x+1)\right)$$ $$\lambda(x)=\underset{p\to \infty }{\text{limit}}A_p(x)=\gamma x+\log (\Gamma (x+1))$$ $$\lambda'(x)=\gamma+\psi ^{(0)}(x+1)=H_x$$

Edit

Otherwise, $$\frac xn-\log\left(1+\frac xn\right)=\sum_{k=2}^\infty (-1)^k\,\frac{ x^k}{k\ n^k}$$ Assuming that we can intergchange the order of summation $$\sum_{n=1}^\infty (-1)^k\,\frac{ x^k}{k\ n^k}=(-1)^k\,\frac{ x^k \zeta (k)}{k}$$ $$\sum_{k=1}^\infty (-1)^k\,\frac{ x^k \zeta (k)}{k}=\gamma x+\log (\Gamma (x+1))$$

If you differentiate deirectly $$\frac d {dx} A_p(x)=\sum_{n=1}^p\left(\frac 1n- \frac{1}{n+x}\right)=H_p+H_x-H_{p+x}$$ Using the asymptotics of harmonic numbers $$H_q=\log (q)+\gamma +\frac{1}{2 q}-\frac{1}{12 q^2}+O\left(\frac{1}{q^4}\right)$$ use it twice and continue with Taylor series to obtain $$\frac d {dx} A_p(x)=H_x-\frac{x}{p}+\frac{x (x+1)}{2 p^2}+O\left(\frac{1}{p^3}\right)$$