The summation in the image represents:

$\sum_{n=1}^{\infty} n \cdot \sin\left(\frac{\pi}{2^n}\right)$

Solution:

To evaluate this series, we analyze each component. Here's the approach:

1. Examine $\sin\left(\frac{\pi}{2^n}\right)$:

   • For large $n$, $\frac{\pi}{2^n} \to 0$, and we can use the approximation $\sin(x) \approx x$ for small $x$. Hence: $\sin\left(\frac{\pi}{2^n}\right) \approx \frac{\pi}{2^n}.$

2. Approximate the series: Substituting the approximation for $\sin$, the series becomes: $\sum_{n=1}^\infty n \cdot \frac{\pi}{2^n} = \pi \sum_{n=1}^\infty \frac{n}{2^n}.$

3. Simplify $\sum_{n=1}^\infty \frac{n}{2^n}$: The summation $\sum_{n=1}^\infty \frac{n}{2^n}$ is a known result. Using the formula: $\sum_{n=1}^\infty \frac{n}{r^n} = \frac{r}{(r-1)^2}, \quad \text{for } r > 1,$ with $r = 2$, we have: $\sum_{n=1}^\infty \frac{n}{2^n} = \frac{2}{(2-1)^2} = 2.$

4. Final result: Substituting back, the series evaluates to: $\pi \cdot 2 = 2\pi.$

Conclusion:

The value of the series is: $\boxed{2\pi}$

Would you like more details or explanations about any of the steps?

Here are 5 follow-up questions:

1. How is the approximation $\sin(x) \approx x$ derived and when is it valid?
2. Can this series be solved exactly without approximation?
3. What are the properties of summations involving geometric or arithmetic progressions?
4. How does the convergence of a series depend on the terms?
5. What is the derivation of $\sum_{n=1}^\infty \frac{n}{r^n}$?

Tip: Approximations like $\sin(x) \approx x$ are powerful tools in series calculations, especially when terms approach zero.