A Combo-Limit+Sum

Let $f = \ln(1+x)$ and let $f_n$ be the sum of the first $n$ terms of the maclaurin expansion of $f$ . Do note that when I use $f^{(n)}$ , it means the nth derivative of $f$ , and when I use $f^n$ , it means the nth exponent of $f$ . The question requires for $\displaystyle J(n) = \lim_{x \rightarrow 0} \frac{1}{f^n}-\frac{1}{f_n^n} = \lim_{x \rightarrow 0} j(n,x)$ .

$\begin{aligned} j(n,x) &= \frac{1}{f^n}-\frac{1}{f_n^n}=\frac{f_n^n-f^n}{f^nf_n^n}=\left(f_n-f\right)\left(\sum_{j=0}^{n-1}f_n^{j-n}\ f^{-1-j}\right) \\ &= -\left(\frac{f^{(n+1)}(a)}{(n+1)!}x^{n+1}\right) \left(\sum_{j=1}^n\frac{1}{\ln(1+x)^j f_n^{n+1-j}}\right) \end{aligned}$

Where $a$ is some real, $0 \le a \le x$ . I used $f - f_n = \frac{f^{(n+1)}(a)}{(n+1)!}x^{n+1}$ that follows from Taylor's Theorem , the Lagrange form of the remainder .

Hence

$\begin{aligned} \lim_{x \rightarrow 0} j(n,x) &= \lim_{x \rightarrow 0}-\frac{f^{(n+1)}(a)}{(n+1)!}x^{n+1}\left(\sum_{j=1}^n\frac{1}{\ln(1+x)^jf_n^{n+1-j}}\right) \\ &=- \frac{f^{(n+1)}(0)}{(n+1)!} \lim_{x \rightarrow 0} \sum_{j=1}^n\frac{x^{n+1}}{\ln(1+x)^jf_n^{n+1-j}} \\ &= -\frac{f^{(n+1)}(0)}{(n+1)!} \lim_{x \rightarrow 0} \sum_{j=1}^n\frac{x^{n+1}}{\ln(1+x)^{n+1}} \\ &= -\frac{n f^{(n+1)}(0)}{(n+1)!} \lim_{x \rightarrow 0} \frac{x^{n+1}}{\ln(1+x)^{n+1}} \\ &=- \frac{n f^{(n+1)}(0)}{(n+1)!} 1^{n+1} \\ &= \left(-1\right)^{n+1}\cdot\frac{n}{n+1} \end{aligned}$

First line to second I used $0 \le a \le x$ . From second to third I used how $f_n$ behaves asymptotically the same as $\ln(1+x)$ when $x$ is close to $0$ . Fourth to fifth I used L'Hopital .

Now the main problem has been solved. The rest follows suit. (See comments for details)

Let $p_n(x) = \frac{-x}{2} + \frac{x^2}{3} - \frac{-x^3}{4} + ... + \frac{(-1)^{n+1}x^{n-1}}{n}$ , such that, $\frac{1}{(x(p_n(x)))^n}= RHS$ .

$J(n) = \lim_{x\to0} \left( \frac{1}{\left(xp_n(x) + \frac{(-1)^{n+2}x^{n+1}}{n+1} + x^{n+2}q_n(x)\right)^n} - \frac{1}{(x(p_n(x)))^n} \right)$ , where $q_n(x)$ is a polynomial. $J(n) = \lim_{x\to0} \frac{1}{x^n - x^{n+1}P_n(x) + nx^{n-1} \frac{(-1)^{n+2}x^{n+1}}{n+1} - x^{2n+1}Q_n(x)} - \frac{1}{x^n - x^{n+1}P_n(x)}$ , for appropriate polynomials $P_n(x), Q_n(x)$ . $J(n) = \lim_{x\to0} \frac{1}{x^n} \left( (1 + (xP_n(x) - n \frac{(-1)^{n+2}x^{n}}{n+1} + x^{n+1}Q_n(x) ) + (xP_n(x) + ... )^2 + ...) - (1 + (xP_n(x)) + (xP_n(x))^2 + ...) \right)$ . inside the brackets, everything of order $x^n$ and below cancels, except $- n \frac{(-1)^{n+2}x^{n}}{n+1}$ .

Essentially, the lowest order difference between the expansion of $\ln(1+x)$ and $xp_n(x)$ is $\frac{(-1)^{n+2}x^{n+1}}{n+1}$ . This only affects the expansion at order $x^{2n}$ when taking it to the power of $n$ . Hence, as there is a common factor of $x^n$ , this component is the difference between the reciprocals of the expansions at order $0$ . All higher orders are ignored in the limit.

Hence, $J(n) = n \frac{(-1)^{n+1}}{n+1}$ .

$A = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(n+1)} = \lim_{x\to -1} \sum_{n=1}^{\infty} \frac{x^n}{n(n+1)} = \lim_{x\to -1} \int \int \left( \sum_{n=0}^{\infty} x^n \right) dx^2 = \lim_{x\to -1} \int \int \frac{1}{1-x} dx^2 = \lim_{x\to -1} \int -\log(1-x) dx$ So, $A = \lim_{x\to -1} x - (x-1)\log(1-x) = -1 + 2\log(2)$ .

$R = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{(n+1)!} = \lim_{x\to -1} \sum_{n=1}^{\infty} \frac{x^{n+1}n}{(n+1)!} = \lim_{x\to -1} \sum_{n=1}^{\infty} x^2 (\frac{nx^{n-1}}{(n+1)!} = \lim_{x\to -1} x^2 (\frac{d}{dx} \frac{e^x}{x} - \frac{1}{x})$ .

So $R= \lim_{x\to -1} x^2 (\frac{e^x}{x} - \frac{e^x}{x^2} + \frac{1}{x^2} ) = \lim_{x\to -1} xe^x - e^x + 1 = 1 - \frac{2}{e}$ .

Hence, $A + R = 2\log(2) - \frac{2}{e}$ .