\sum\limits_{n = 1}^\infty  {\left( {\dfrac{{\sum\limits_{r = 1}^n {\left( r \right)} }}{{{2^n}}}} \right)} 

To find the value of the infinite series 

\[ \sum\limits_{n = 1}^\infty  \left( \dfrac{{\sum\limits_{r = 1}^n {r} }}{{2^n}} \right), \]

we need to evaluate the inner sum and then the outer sum.

1. **Evaluate the inner sum:**

   The inner sum \( \sum\limits_{r = 1}^n r \) is the sum of the first \( n \) natural numbers, which is given by the formula:

   \[ \sum\limits_{r = 1}^n r = \frac{n(n+1)}{2}. \]

2. **Substitute the inner sum into the outer sum:**

   Substitute \( \sum\limits_{r = 1}^n r = \frac{n(n+1)}{2} \) into the outer sum:

   \[ \sum\limits_{n = 1}^\infty  \left( \dfrac{\frac{n(n+1)}{2}}{2^n} \right) = \sum\limits_{n = 1}^\infty  \left( \frac{n(n+1)}{2 \cdot 2^n} \right) = \sum\limits_{n = 1}^\infty  \left( \frac{n(n+1)}{2^{n+1}} \right). \]

3. **Separate the terms:**

   We can separate the fraction into two sums:

   \[ \sum\limits_{n = 1}^\infty  \left( \frac{n(n+1)}{2^{n+1}} \right) = \sum\limits_{n = 1}^\infty  \left( \frac{n^2 + n}{2^{n+1}} \right) = \sum\limits_{n = 1}^\infty  \left( \frac{n^2}{2^{n+1}} + \frac{n}{2^{n+1}} \right). \]

4. **Split into two sums:**

   Split the sum into two separate sums:

   \[ \sum\limits_{n = 1}^\infty  \left( \frac{n^2}{2^{n+1}} \right) + \sum\limits_{n = 1}^\infty  \left( \frac{n}{2^{n+1}} \right). \]

5. **Evaluate each sum separately:**

   To evaluate these sums, we can use known series results.

   - **First sum:** \( \sum\limits_{n = 1}^\infty \frac{n}{2^n} \)

     This is a known geometric series sum that can be evaluated using the formula for the sum of an arithmetico-geometric series:

     \[ \sum\limits_{n = 1}^\infty \frac{n}{2^n} = 2. \]

   - **Second sum:** \( \sum\limits_{n = 1}^\infty \frac{n^2}{2^n} \)

     Similarly, this sum can be evaluated using advanced series techniques or known results:

     \[ \sum\limits_{n = 1}^\infty \frac{n^2}{2^n} = 6. \]

6. **Combine the results:**

   Now combine the results, remembering the factor \( \frac{1}{2} \) in each term:

   \[ \sum\limits_{n = 1}^\infty  \left( \frac{n^2}{2^{n+1}} \right) = \frac{1}{2} \sum\limits_{n = 1}^\infty  \left( \frac{n^2}{2^n} \right) = \frac{1}{2} \cdot 6 = 3. \]

   \[ \sum\limits_{n = 1}^\infty  \left( \frac{n}{2^{n+1}} \right) = \frac{1}{2} \sum\limits_{n = 1}^\infty  \left( \frac{n}{2^n} \right) = \frac{1}{2} \cdot 2 = 1. \]

7. **Final result:**

   Adding these together gives:

   \[ 3 + 1 = 4. \]

Therefore, the value of the series 

\[ \sum\limits_{n = 1}^\infty  \left( \dfrac{{\sum\limits_{r = 1}^n {r} }}{{2^n}} \right) \]

is \( 4 \).

If you have any questions or need further clarification, please let me know.

Here are some additional questions you might find interesting:
1. How do you derive the formula for the sum of the first \( n \) natural numbers?
2. What is the arithmetico-geometric series and how is it used?
3. Can you explain the method to find the sum of geometric series?
4. What other techniques can be used to evaluate infinite series?
5. How do we verify the convergence of a given infinite series?

**Tip:** Knowing common series sums and their derivations can be extremely helpful in solving more complex series problems.

Learn how to evaluate the infinite series sum \( \sum\limits_{n = 1}^\infty  \left( \dfrac{{\sum\limits_{r = 1}^n {r} }}{{2^n}} \right) \) step-by-step. Understand the inner and outer sums involved, including the use of geometric series sum formulas. Suitable for advanced mathematics students.

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