Interesting Limit (10)

Note that $T_n \; = \; \sum_{j=n}^{2n-1} \frac{2}{2j+1} \; = \; 2\left(\sum_{j=0}^{2n-1}\frac{1}{2j+1} - \sum_{j=0}^{n-1}\frac{1}{2j+1}\right) \; = \; 2\left(\sum_{j=1}^{4n}\frac{1}{j} - \sum_{j=1}^{2n}\frac{1}{2j} - \sum_{j=1}^{2n}\frac{1}{j} + \sum_{j=1}^n \frac{1}{2j}\right) \; = \; 2H_{2n} - 3H_{2n} + H_n$ and so $T_n = \ln2 + 2S_{4n} - 3S_{2n} + S_n$ , where $S_n = H_n - \ln n$ . Since $S_n \; = \; \gamma + \frac{1}{2n} - \frac{1}{12n^2} + o\big(n^{-2}\big) \hspace{2cm} n \to \infty$ we deduce that $2S_{4n} - 3S_{2n} + S_n \; = \; -\tfrac{1}{32}n^{-2} + o\big(n^{-2}\big) \hspace{2cm} n \to \infty$ and hence $\lim_{n \to \infty} n^2\big[\ln2 - T_n\big] \; = \; \tfrac{1}{32}$ so that $A = \tfrac{1}{32}$ and $B = \ln2$ . Thus $e^{\frac{B}{A}} = 2^{32}$ , making the answer $\sqrt[8]{2^{32}} = 2^4 = \boxed{16}$ .

Let us first see that $\displaystyle \lim_{n\to\infty}\sum_{r=1}^{n}\frac{2}{2n+2r-1} = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\frac{1}{1+\frac{r}{n}-\frac{1}{2n}}=\int_{0}^{1}\frac{1}{1+x}dx = \ln(2)$

Hence $B=\ln(2)$ otherwise the limit would tend to infinity . But $B=\ln(2)$ ensures that we have a $\frac{0}{0}$ form.

Now we have:-

$\displaystyle \lim_{n\to\infty}\frac{\left(\ln(2) -\sum_{r=1}^{n}\frac{2}{2n+2r-1}\right)}{\frac{1}{n^{2}}}$

As $\frac{1}{n^{2}}$ is monotone and tends to 0 we can apply Stolz-Cesaro Theorem

Hence we have $\displaystyle \lim_{n\to\infty}\frac{\left(\ln(2) -\sum_{r=1}^{n}\frac{2}{2n+2r-1}\right)}{\frac{1}{n^{2}}} = \lim_{n\to\infty} \frac{2\left(\sum_{r=1}^{n+1}\frac{1}{2(n+1)+2r-1} - \sum_{r=1}^{n}\frac{1}{2n+2r-1}\right)}{\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}}$

Simplifying the above we have:-

$\displaystyle \lim_{n\to\infty}\frac{2\left(\frac{1}{4n+3}+\sum_{r=1}^{n}\left(\frac{1}{2n+(2r+1)}-\frac{1}{2n+(2r-1)}\right)\right)}{\frac{2n+1}{n^{2}(n+1)^{2}}}$

Simplifying further we have:-

$\displaystyle \lim_{n\to\infty}\frac{2\left(\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{1}{2n+1}\right)}{\frac{2n+1}{n^{2}(n+1)^{2}}}$

Simplifying further we have:-

$\displaystyle \lim_{n\to\infty}\frac{2\left((8n+4)(2n+1)-(4n+1)(4n+3)\right)}{(4n+1)(4n+3)(2n+1)^{2}}\cdot (n^{2}(n+1)^{2})=\lim_{n\to\infty}2\cdot\frac{16n^{2}+16n+4-16n^{2}-16n-3}{(4n+3)(4n+3)(2n+1)^{2}}\cdot n^{2}\cdot (n+1)^{2}= \lim_{n\to\infty}\frac{2n^{2}\cdot (n+1)^{2}}{(4n+1)(4n+3)(2n+1)(2n+1)} =\frac{2}{4^{3}}= \frac{1}{32}$

Hence $\large (\text{exp}(\frac{B}{A}))^{\frac{1}{8}}=\text{exp}(\ln(16)) =16$

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

First note that $\frac{2}{2n+1} + \frac{2}{2n+3} + \cdots + \frac{2}{4n-1} = \sum_{k=1}^n \frac{2}{2n+2k-1} = \frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k-1/2}{n}} \xrightarrow{n\to\infty} \int_0^1 \frac{1}{1+x}\,dx = \ln 2,$ so that we must have $B = \ln 2$ . Also, using the integral form for $B$ and setting $f(x) = \frac{1}{1+x}$ , we have $B - \frac{1}{n} \sum_{k=1}^n f\left(\frac{k-1/2}{n}\right) = \sum_{k=1}^n \int_{(k-1)/n}^{k/n} \left(f(x) - f\left(\frac{k-1/2}{n}\right)\right)\,dx$

To evaluate this, we temporarily fix $k,n$ and use Taylor's theorem twice to write $\begin{aligned} \int_{(k-1)/n}^{(k-1/2)/n} \left(f(x) - f\left(\frac{k-1/2}{n}\right)\right)\,dx &= -\frac{1}{2}f'\left(\frac{k-1/2}{n}\right)\cdot \left(\frac{1}{2n}\right)^2 + \frac{1}{6} f''(\eta_1)\left(\frac{1}{2n}\right)^3 \\[1ex] \int_{(k-1/2)/n}^{k/n} \left(f(x) - f\left(\frac{k-1/2}{n}\right)\right)\,dx &= +\frac{1}{2}f'\left(\frac{k-1/2}{n}\right)\cdot \left(\frac{1}{2n}\right)^2 + \frac{1}{6} f''(\eta_2)\left(\frac{1}{2n}\right)^3 \end{aligned}$ for some $\frac{k-1}{n} < \eta_1 < \frac{k-1/2}{n} < \eta_2 < \frac{k}{n}$ . Then adding these two gives $\int_{(k-1)/n}^{k/n} \left(f(x) - f\left(\frac{k-1/2}{n}\right)\right)\,dx = \frac{1}{24n^3} \frac{f''(\eta_1) + f''(\eta_2)}{2} = \frac{1}{24n^3}f''(x_{k,n})$ for some $x_{k,n} \in (\eta_1, \eta_2) \subset \left(\frac{k-1}{n},\frac{k}{n}\right)$ .

Going back to the main problem, we have $\begin{aligned} A &= \lim_{n\to\infty} n^2\left(B - \left(\frac{2}{2n+1} + \frac{2}{2n+3} + \cdots + \frac{2}{4n-1}\right)\right) \\ &= \lim_{n\to\infty} n^2 \sum_{k=1}^n \frac{1}{24n^3}f''(x_{k,n}) \\ &= \frac{1}{24} \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^n f''(x_{k,n}) \\ &= \frac{1}{24} \int_0^1 f''(x)\,dx \\ &= \frac{1}{24} \left(f'(1) - f'(0)\right) \\ &= \frac{1}{24} \left(-\frac{1}{(1+1)^2} + \frac{1}{(1+0)^2}\right) \\ &= \frac{1}{32} \end{aligned}$

Therefore $\large \sqrt[8]{e^{\frac{B}{A}}} = e^{\frac{32\ln 2}{8}} = 2^4 = \boxed{16}$