Beautiful Limit

$\begin{aligned} L&=\lim_{n\to\infty}\prod\limits_{r=1}^n\sqrt[r]{1+\left(\frac{r}{n}\right)^{168}}\\ I=\ln(L)&=\lim_{n\to\infty}\sum\limits_{r=1}^n\frac{1}{r}\ln\left(1+\left(\frac{r}{n}\right)^{168}\right)\\ &=\lim_{n\to\infty}\sum\limits_{r=1}^n\frac 1n\frac{n}{r}\ln\left(1+\left(\frac{r}{n}\right)^{168}\right)=\int\limits_0^1 \frac 1x\ln\left(1+x^{168}\right)\,\mathrm{d}x\\ &\overset{u=x^{168}}=\frac{1}{168}\int\limits_0^1 \frac 1u\ln\left(1+u\right)\,\mathrm{d}u\\ &=\frac{1}{168}\int\limits_0^1 \frac 1u\sum\limits_{k=1}^\infty (-1)^{k+1}\frac{u^k}{k}\,\mathrm{d}u\\ &=\left.\frac{1}{168} \sum\limits_{k=1}^\infty (-1)^{k+1}\frac{u^k}{k^2}\right|_0^1=\frac{1}{168} \sum\limits_{k=1}^\infty (-1)^{k+1}\frac{1}{k^2}\\ &=\frac{1}{168}\cdot\frac{\pi^2}{12}~~~~~~~~~~~~~~~~\color{#3D99F6} \text{see below}\\ &=\frac{\pi^2}{2016}\\\\ L&=\mathrm e^I \end{aligned}$

So $a+b=\boxed{2018}$

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Mostly you can write a periodical function $f(x)$ with period L as a fourier series:

$\begin{aligned} f(x)&=\frac{a_0}{\sqrt{L}}+\sqrt{\frac 2L}\sum_{n=1}^\infty\left[a_n\cos(k_nx)+b_n\sin(k_nx)\right]\\ k_n&=2\pi n/L\\\\ a_0&=\frac{1}{\sqrt{L}}\int\limits_0^Lf(x)\,\mathrm{d}x\\ a_n&=\sqrt{\frac{2}{L}}\int\limits_0^Lf(x)\cos(k_nx) \,\mathrm{d}x\\ b_n&=\sqrt{\frac{2}{L}}\int\limits_0^Lf(x)\sin(k_nx) \,\mathrm{d}x \end{aligned}$

Now consider:

$\begin{aligned} f(x)&=x^2~~~~~,~-\pi\leq x\leq\pi~~\text{(and periodicaly continued)}\\ L&=2\pi\\\\ a_0&=\frac{1}{\sqrt{2\pi}}\frac 23\pi^3\\ a_k&=\frac{1}{\sqrt{\pi}}\frac{4\pi(-1)^k}{k^2}\\ \Rightarrow f(0)=0&=\frac 13\pi^2+4\sum\limits_{k=1}^\infty\frac{(-1)^k}{k^2}\\\\ \Rightarrow \sum\limits_{k=1}^\infty\frac{(-1)^{k+1}}{k^2}&=-\sum\limits_{k=1}^\infty\frac{(-1)^k}{k^2}=\frac{\pi^2}{12} \end{aligned}$