lim
n
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∞

n
∑
k
=
1

(
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sin
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π
2
k
−
(
cos
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π
2
k
−
(
sin
)
(
π
2
(
k
+
2
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+
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cos
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π
2
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k
+
2
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)

To solve the given limit involving a sum, we need to analyze the expression carefully. The expression given is:

\[
\lim_{n \to \infty} \sum_{k=1}^{n} \left( \sin\left(\frac{\pi}{2}k\right) - \cos\left(\frac{\pi}{2}k\right) - \left(\sin\left(\frac{\pi}{2}(k+2)\right) + \cos\left(\frac{\pi}{2}(k+2)\right)\right)\right)
\]

This expression can be broken down into two parts:

1. \( \sin\left(\frac{\pi}{2}k\right) - \cos\left(\frac{\pi}{2}k\right) \)
2. \( \sin\left(\frac{\pi}{2}(k+2)\right) + \cos\left(\frac{\pi}{2}(k+2)\right) \)

The terms involving \( k \) and \( k+2 \) suggest that this might telescope, which means most terms in the sum will cancel out, leaving only a few terms at the beginning and end.

### Step-by-step Solution

#### 1. Evaluate the trigonometric expressions:

Trigonometric functions like \(\sin\) and \(\cos\) with arguments of \(\frac{\pi}{2}k\) have periodic behavior:

- \( \sin\left(\frac{\pi}{2}k\right) \) will cycle through the values \(0, 1, 0, -1\) as \(k\) changes.
- \( \cos\left(\frac{\pi}{2}k\right) \) will cycle through the values \(1, 0, -1, 0\) as \(k\) changes.

#### 2. Use these cycles to find specific values:

For \(k = 1\) to \(k = 4\), let's see how these functions evaluate:

- **\(k = 1\):** \(\sin\left(\frac{\pi}{2} \cdot 1\right) = 1\), \(\cos\left(\frac{\pi}{2} \cdot 1\right) = 0\)
- **\(k = 2\):** \(\sin\left(\frac{\pi}{2} \cdot 2\right) = 0\), \(\cos\left(\frac{\pi}{2} \cdot 2\right) = -1\)
- **\(k = 3\):** \(\sin\left(\frac{\pi}{2} \cdot 3\right) = -1\), \(\cos\left(\frac{\pi}{2} \cdot 3\right) = 0\)
- **\(k = 4\):** \(\sin\left(\frac{\pi}{2} \cdot 4\right) = 0\), \(\cos\left(\frac{\pi}{2} \cdot 4\right) = 1\)

Similarly, find for \(k+2\).

#### 3. Find a pattern for the sum:

The expression telescopes:

\[
\sin\left(\frac{\pi}{2}k\right) - \cos\left(\frac{\pi}{2}k\right) - \sin\left(\frac{\pi}{2}(k+2)\right) - \cos\left(\frac{\pi}{2}(k+2)\right)
\]

Since the sine and cosine terms cancel out due to their periodicity, evaluate for initial terms and find a pattern. Because of the periodic nature:

The sum of the remaining terms is constant over each full cycle of 4 steps. Let's calculate one full cycle:

- For \(k=1\) to \(k=4\):
  - Cycle of terms (before and after simplification):
  \[
  \begin{align*}
  & (1-0) - (0+(-1)) = 2,\\
  & (0+1) - ((-1)+0) = 2,\\
  & (-1-0) - (0+1) = -2,\\
  & (0-(-1)) - (1+0) = -2
  \end{align*}
  \]

Thus the sequence telescopes to zero, showing the finite sum \(=0\) over each cycle.

#### Conclusion:

Since each group of four terms adds to zero, and because \(n\) approaches infinity with cycles completing, the entire sum also approaches zero.

\[
\boxed{0}
\]

Feel free to ask for more details, or let me know if you have any questions!

### Additional Questions to Consider:

1. What is a telescoping series, and how does it simplify complex sums?
2. How do trigonometric identities assist in finding patterns in sequences?
3. What is the significance of periodic functions in series calculations?
4. How can evaluating initial terms of a sequence help predict the behavior of a sum?
5. Can a telescoping sequence converge to a non-zero value? Under what conditions?
6. How does the behavior of \(\sin\) and \(\cos\) over their periods affect sums?
7. What are other common series where telescoping is used?
8. How can we rigorously prove the cancellation pattern observed?

**Tip:** Remember that identifying patterns and periodicity in sequences can greatly simplify complex summations and limit evaluations.

Explore how to solve the limit of a telescoping series involving trigonometric functions. Understand the behavior of sine and cosine functions over their periods and how it affects the sum of infinite terms. Suitable for advanced undergraduate mathematics.

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n ∑ k

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