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 $$\lambda(1)=\gamma$$
Best Answer
$$\frac{d}{dx}\sum_{n=1}^\infty\left(\frac xn-\ln\left(1+\frac xn\right)\right)=\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+x}\right)$$
$$=\lim_{N\to \infty}\sum_{n=1}^N\left(\frac{1}{n}-\frac{1}{n+x}\right)$$
$$=\lim_{N\to \infty}\left[\sum_{n=1}^N\frac{1}{n}-\sum_{n=1}^N\frac{1}{n+x}\right]$$
$$=\lim_{N\to \infty}\left[\sum_{n=1}^N\frac{1}{n}-\left(\underbrace{\sum_{n=1}^{N-x}\frac{1}{n+x}}_{n\to n-x}+\underbrace{\sum_{n=N-x+1}^{N}\frac{1}{n+x}}_{n\to N-n}\right)\right]$$
$$=\lim_{N\to \infty}\left[\sum_{n=1}^N\frac{1}{n}-\sum_{n=x+1}^{N}\frac{1}{n}-\sum_{n=0}^{x-1}\frac{1}{N+x-n}\right]$$
$$=\lim_{N\to \infty}\left(\sum_{n=1}^N\frac{1}{n}-\sum_{n=x+1}^{N}\frac{1}{n}\right)-\lim_{N\to \infty}\sum_{n=0}^{x-1}\frac{1}{N+x-n}$$
$$=\lim_{N\to \infty}\left(\sum_{n=1}^x\frac{1}{n}\right)-\lim_{N\to \infty}\sum_{n=0}^{x-1}\frac{1}{N+x-n}$$
$$=\sum_{n=1}^x\frac{1}{n}-0$$
$$=H_x$$