To solve the limit
\[
\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{\sin\left(\frac{(n+r)\pi}{4n}\right)} \cdot \frac{\pi}{n},
\]
we will follow these steps:
### Step 1: Rewrite the expression inside the limit
We start by rewriting the sine term in the summation:
\[
\sin\left(\frac{(n+r)\pi}{4n}\right) = \sin\left(\frac{n\pi}{4n} + \frac{r\pi}{4n}\right) = \sin\left(\frac{\pi}{4} + \frac{r\pi}{4n}\right).
\]
### Step 2: Use the sine addition formula
Using the sine addition formula, we have:
\[
\sin\left(\frac{\pi}{4} + x\right) = \sin\left(\frac{\pi}{4}\right)\cos(x) + \cos\left(\frac{\pi}{4}\right)\sin(x),
\]
where \(x = \frac{r\pi}{4n}\). Thus,
\[
\sin\left(\frac{\pi}{4} + \frac{r\pi}{4n}\right) = \frac{\sqrt{2}}{2}\cos\left(\frac{r\pi}{4n}\right) + \frac{\sqrt{2}}{2}\sin\left(\frac{r\pi}{4n}\right).
\]
### Step 3: Simplify the summation
Now we can rewrite the summation:
\[
\sum_{r=1}^{n} \frac{1}{\sin\left(\frac{\pi}{4} + \frac{r\pi}{4n}\right)} \cdot \frac{\pi}{n}.
\]
As \(n \to \infty\), \(\frac{r}{n} \to x\) where \(x\) varies from \(0\) to \(1\). Thus, we can replace the summation with an integral:
\[
\int_0^1 \frac{\pi}{\sin\left(\frac{\pi}{4} + \frac{\pi}{4}x\right)} \, dx.
\]
### Step 4: Evaluate the integral
Now we need to evaluate the integral:
\[
\pi \int_0^1 \frac{1}{\sin\left(\frac{\pi}{4} + \frac{\pi}{4}x\right)} \, dx.
\]
Using the substitution \(u = \frac{\pi}{4} + \frac{\pi}{4}x\), we have \(du = \frac{\pi}{4}dx\) or \(dx = \frac{4}{\pi}du\). The limits change accordingly:
- When \(x = 0\), \(u = \frac{\pi}{4}\).
- When \(x = 1\), \(u = \frac{\pi}{2}\).
Thus, the integral becomes:
\[
\pi \cdot \frac{4}{\pi} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin(u)} \, du = 4 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \csc(u) \, du.
\]
### Step 5: Evaluate the integral of cosecant
The integral of \(\csc(u)\) is given by:
\[
\int \csc(u) \, du = \ln\left|\tan\left(\frac{u}{2}\right)\right| + C.
\]
Thus, we evaluate:
\[
4 \left[ \ln\left|\tan\left(\frac{u}{2}\right)\right| \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}.
\]
Calculating the limits:
- At \(u = \frac{\pi}{2}\), \(\tan\left(\frac{\pi}{4}\right) = 1\), so \(\ln(1) = 0\).
- At \(u = \frac{\pi}{4}\), \(\tan\left(\frac{\pi}{8}\right)\).
Thus, we have:
\[
4 \left(0 - \ln\left(\tan\left(\frac{\pi}{8}\right)\right)\right) = -4 \ln\left(\tan\left(\frac{\pi}{8}\right)\right).
\]
### Final Result
Using the identity \(\tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1\), we find:
\[
\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{\sin\left(\frac{(n+r)\pi}{4n}\right)} \cdot \frac{\pi}{n} = -4 \ln(\sqrt{2} - 1).
\]
