The image you provided shows a summation problem with a series, and the specific equation to solve seems to be related to a summation of terms, likely approaching a limit as $n \to \infty$.

Here is a breakdown of what the image appears to contain:

1. First Line (Limit of the series): $\lim_{n \to \infty} \left( \frac{1}{2} + \frac{8}{2^2} + \frac{5}{2^3} + \dots + \frac{2n-1}{2^n} \right)$

2. Second Line (Summation definition): $S_n = \sum_{k=1}^{n} \frac{2k-1}{2^k}$

This is a summation where each term involves both the index $k$ and a geometric progression in the denominator.

Solution Strategy:

The sum can be split into two parts, for easier calculation:

$S_n = \sum_{k=1}^{n} \frac{2k}{2^k} - \sum_{k=1}^{n} \frac{1}{2^k}$

Step 1: Simplify each summation

• The first sum: $\sum_{k=1}^{n} \frac{2k}{2^k}$

• The second sum: $\sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}$

Let me compute the rest and evaluate this sum.The two parts of the summation simplify as follows:

1. The first part, $\sum_{k=1}^{\infty} \frac{2k}{2^k}$, converges to 4.
2. The second part, $\sum_{k=1}^{\infty} \frac{1}{2^k}$, converges to 1.

Therefore, the total sum $S_n$ for $n \to \infty$ is:

$S_{\infty} = 4 - 1 = 3$

Thus, the result of the series is 3.

Would you like more details on any of the steps? Here are some related questions to explore:

1. What is the general formula for summing geometric series?
2. How does the convergence of series behave as $n \to \infty$?
3. Can this approach be applied to other types of series?
4. What role does the ratio of the geometric series play in the convergence?
5. How would the result change if the terms of the series were slightly altered?

Tip: When dealing with infinite sums, splitting the series into manageable parts often simplifies the evaluation.