Derivative and Infinite Sum

Note:

$\frac{d^n}{dx^n}\bigg(\frac{\log{x}}{x}\bigg)\Biggr|_{\substack{x=1}}=\frac{d^n}{dx^n}\bigg(\frac{\log{(1+x)}}{1+x}\bigg)\Biggr|_{\substack{x=0}}$

$\frac{\log{(1+x)}}{1+x} = \bigg(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots\bigg)\bigg(1 - x + x^2 - x^3 - \dots\bigg) \dots (2)$

$\text{Co-efficient of } x^n \text{ in above equation}= (-1)^{n+1}\bigg(1 + \frac{1}{2} + \frac{1}{3} + \dots \frac{1}{n}\bigg)$

For evaluating the n-th derivative of a polynomial (or more appropriately, power series here) at x=0, only information of the co-efficient of $x^n$ is required as all the lower powers will disappear due to repeated differentiation and higher powers will vanish on putting x=0. (This is why the transformation in first equation was done.)

Formally:

$\text{If }f(x) = a_0 + a_1x + a_2x^2 + \dots$

$=\sum\limits_{n=0}^{\infty}a_nx^n$

Then,

$\frac{d^n}{dx^n}f(x)\Biggr|_{x=0} = n!\cdot a_n$

Thus, we have:

$g(n) = (-1)^{n+1}\bigg(1 + \frac{1}{2} + \frac{1}{3} + \dots \frac{1}{n}\bigg)$

Now, note that for $k\in\mathbb{N}$ , the following holds:

$g(2k)+g(2k+1) = -\bigg(1+\frac{1}{2}+\dots\frac{1}{2k}\bigg) + \bigg(1+\frac{1}{2}+\dots\frac{1}{2k+1}\bigg)=\frac{1}{2k+1}$

Finally,

$\sum\limits_{k=2m}^{4m+1}g(k) = g(2m) + g(2m+1) + g(2m+2) + g(2m+3) + \dots g(4m) + g(4m+1)$

$= \frac{1}{2m+1} + \frac{1}{2m+3} + \dots \frac{1}{4m+1}$

$= \sum_{r=0}^{m}\frac{1}{2(m+r)+1}$

$\therefore \lim\limits_{m \to \infty}\sum\limits_{k=2m}^{4m+1}g(k) = \lim\limits_{m \to \infty}\sum\limits_{r=0}^{m}\frac{1}{2(m+r)+1}$

$=\lim\limits_{m \to \infty}\frac{1}{m}\sum\limits_{r=0}^{m}\dfrac{1}{2+\dfrac{2r+1}{m}}$

If $\frac{r+\frac{1}{2}}{m}=x\text{, then }\frac{1}{m}=dx$

$\therefore \lim\limits_{m \to \infty}\sum\limits_{k=2m}^{4m+1}g(k) = \int\limits_{0}^{1}\frac{1}{2+2x}dx$

$\boxed{=\frac{1}{2}\log{2}}$