Limit of Sine sum

Let us rewrite the series as :- $\Large \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n}\frac{\pi}{1+\frac{r}{n}} \sin(\frac{1}{n}\frac{\pi}{1+\frac{r}{n}})\frac{1}{(\frac{1}{n})(\frac{\pi}{1+\frac{1}{n}})}$

Now the sequence $\large \frac{1}{n}\frac{\pi}{1+\frac{r}{n}}$ tends to $0$ as $n\to\infty$

So let us call it $\large \theta =\frac{1}{n}\frac{\pi}{1+\frac{r}{n}}$ . Hence $\theta \to 0$ as $n \to \infty$ .

So $\large \lim_{n\to\infty} (\sin(\frac{1}{n}\frac{\pi}{1+\frac{r}{n}}))\frac{1}{(\frac{1}{n}\frac{\pi}{1+\frac{1}{n}})} = \lim_{\theta\to0} \frac{sin(\theta)}{\theta}$ tends to $1$ as $n\to\infty$

So the sum is equivalent to :-

$\large \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n}\frac{\pi}{1+\frac{r}{n}}$

Using Riemann Sums , we have:-

$\Large \int_{0}^{1} \frac{\pi}{1+x} dx = \pi \ln(2)$

The given expression can be rewritten as $\lim_{n\to \infty}\sum_{i=1}^n \dfrac{\sin \frac{π}{n+i}}{\frac{π}{n+i}}\times \dfrac{π}{n+i}=π\lim_{h\to 0, nh=1} \sum_{i=1}^n \dfrac{h}{1+ih}=π\displaystyle \int_0^1 \dfrac{dx}{1+x}=π\ln 2$ . So, $a=π, b=2$ , and $a^b+b^a=π^2+2^π\approx \boxed {18.6945}$ .

Call the limit of sum $\displaystyle L(n)= \lim_{n\to\infty} \sum_{k=1}^n\sin\frac{\pi}{n+k}$ . By Taylor theorem one can easily deduce that for $x>0$ the following inequality holds true. $x-\frac{x^3}{6}< \sin x< x$ and for $x=\frac{\pi}{n+k}$ and $k>0$ we have $\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{\pi}{n+k}-\frac{\pi^3}{6(n+k)^3}\right)< L(n)<\lim_{n\to\infty}\sum_{k=1}^n\frac{\pi}{n+k}$ Also we note that $\frac{1}{2n}\leq \frac{1}{n+k}\leq \frac{1}{n+1}$ as $n+1\leq n+k\leq 2n$ for $1\leq k\leq n$ ( prove by induction ) and hence $\lim_{n\to\infty}\frac{\pi}{8n^3}\sum_{k=1}^n \leq L'(n)\leq \frac{ \pi}{(n+1)^3}\lim_{n\to\infty}\sum_{k=1}^n$ therefore the limit of $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{\pi}{6(n+k)^3}=L'(n)=0$ by squeeze theorem as the $\lim_{n\to\infty}\frac{\pi}{8n^3}\sum_{k=1}^n = \lim_{n\to\infty}\frac{n\pi}{8n^3}=0=\lim_{n\to\infty}\frac{\pi n}{(n+1)^3}$ and hence the original limit (by squeeze theorem) $L(n)=\lim_{n\to\infty}\sum_{k=1}^n\frac{\pi}{n+k}=\int_0^1\frac{\pi}{1+x}dx= \pi\ln 2$ and the required answer is $\pi^2+2^{\pi}\approx 18.6945$ .