Interesting Limit (9)

Relevant wiki: Taylor's Theorem (with Lagrange Remainder)

A necessary requirement for the existence of a limit of the form $\lim_{n\to\infty} n\cdot f(n)$ is that $\lim_{n\to\infty} f(n) = 0$ . Applying it here, we have $B = \lim_{n\to\infty} \sum_{j=1}^n \frac{1}{n+j} = \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n \frac{1}{1+\frac{j}{n}} = \int_0^1 \frac{1}{1+x}\,dx = \ln(2)$

Also, using the form of $B$ as an integral, we may let $f(x) = \frac{1}{1+x}$ to write $\begin{aligned} n\left(B - \frac{1}{n}\sum_{j=1}^n \frac{1}{1+\frac{j}{n}}\right) &= n\sum_{j=1}^n \int_{(j-1)/n}^{j/n} \left(\frac{1}{1+x} - \frac{1}{1+\frac{j}{n}}\right)\,dx \\ &= n\sum_{j=1}^n \int_{(j-1)/n}^{j/n} \left(f(x) - f\left(\frac{j}{n}\right)\right)\,dx \\ &= n\sum_{j=1}^n -\frac{1}{2}f'(x_{j,n})\cdot \frac{1}{n^2} \qquad \text{for some } x_{j,n} \in \left(\frac{j-1}{n}, \frac{j}{n}\right) & \textbf{[see the note]}\\ &= -\frac{1}{2} \cdot \frac{1}{n} \sum_{j=1}^n f'(x_{j,n}) \end{aligned}$

Letting $n\to\infty$ , we find $A = -\frac{1}{2} \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n f'(x_{j,n}) = -\frac{1}{2} \int_0^1 f'(x)\,dx = -\frac{1}{2} \left(f(1) - f(0)\right) = -\frac{1}{2} \left(\frac{1}{1+1} - \frac{1}{1+0}\right) = \frac{1}{4}$

Therefore $e^{B/A} = e^{4\ln(2)} = \boxed{16}$

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Note : The line in my solution I marked to see this note is the only line that isn't straight forward. It is an immediate consequence of Taylor's theorem in the following form: $\text{If } g(x) \text{ has a continuous first derivative, then } \\ \int_a^b g(x)\,dx = g(b)(b-a) - \frac{1}{2}g'(x_1^*)(b-a)^2 = g(a)(b-a) + \frac{1}{2}g'(x_2^*)(b-a)^2 \\ \text{ for some } x_1^*, x_2^* \in(a,b)$

$B=\displaystyle\lim_{n\to\infty}\left(\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{2n}\right)=\ln(2n)-\ln n=\ln2$
$A=\displaystyle\lim_{n\to\infty}n\left(B-\left(\frac 1{n+1}+\frac 1{n+2}+\cdots+\frac 1{2n}\right)\right)$ ... (1)
$=\displaystyle\lim_{2n\to\infty}2n\left(B-\left(\frac 1{2n+1}+\frac 1{2n+2}+\cdots+\frac 1{4n}\right)\right)$
$=\displaystyle\lim_{n\to\infty}n\left(2B-\left(\frac 1{n+0.5}+\frac 1{n+1}+\frac1{n+1.5}+\cdots+\frac 1{2n}\right)\right)$ ... (2)
(1) + (1) - (2) $=A=\displaystyle\lim_{n\to\infty}n\left(\frac 1{n+0.5}-\frac 1{n+1}+\frac1{n+1.5}-\cdots-\frac 1{2n}\right)$
$=\displaystyle\lim_{n\to\infty}n\left(\frac {0.5}{(n+0.5)(n+1)}+\frac {0.5}{(n+1.5)(n+2)}+\cdots+\frac {0.5}{(2n-0.5)(2n)}\right)$
$=0.5\displaystyle\lim_{n\to\infty}n\left(\frac 1{(n+0.5)(n+1)}+\frac 1{(n+1.5)(n+2)}+\cdots+\frac 1{(2n-0.5)(2n)}\right)$
$=0.5\displaystyle\lim_{n\to\infty}n\left(\frac 1{(n+1)^{2}}+\frac 1{(n+2)^2}+\cdots+\frac 1{(2n)^2}\right)$
$=0.5\displaystyle\lim_{n\to\infty}\frac1{n}\left(\frac {n^2}{(n+1)^{2}}+\frac {n^2}{(n+2)^2}+\cdots+\frac {n^2}{(2n)^2}\right)$
$=0.5\displaystyle\int_{1}^{2}\frac{1}{n^2}=\frac{1}{4}$

We get ${e}^{\frac BA}= \boxed{16}$