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The problem is asking to compute the 5th partial sum, \( S_5 \), of the series

\[
\sum_{n=1}^{\infty} \frac{(-1)^n}{2n+1}.
\]

The 5th partial sum, \( S_5 \), is given by:

\[
S_5 = \sum_{n=1}^{5} \frac{(-1)^n}{2n+1}.
\]

This means we need to compute:

\[
S_5 = \frac{(-1)^1}{2(1)+1} + \frac{(-1)^2}{2(2)+1} + \frac{(-1)^3}{2(3)+1} + \frac{(-1)^4}{2(4)+1} + \frac{(-1)^5}{2(5)+1}.
\]

Let's break this down:

\[
S_5 = \frac{-1}{3} + \frac{1}{5} + \frac{-1}{7} + \frac{1}{9} + \frac{-1}{11}.
\]

Now, let's calculate this sum step by step.

\[
S_5 = -\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11}.
\]

I will now compute the exact value.The 5th partial sum, \( S_5 \), is approximately:

\[
S_5 \approx -0.256.
\]

Would you like further details on the calculation or have any questions?

Here are five related questions you might find useful:

1. How does the 5th partial sum compare to the full sum of the series?
2. How can you determine if the infinite series converges?
3. How does the alternating sign in the series affect the sum?
4. What would the 6th partial sum of this series look like?
5. Can this series be expressed in terms of a well-known mathematical constant?

**Tip:** Alternating series like this one often converge conditionally, meaning the terms get smaller and the signs alternate, leading to a convergent sum.

Given the series ∑ (-1)^n / (2n + 1) from n=1 to infinity, find the 5th partial sum, s_5.

Learn how to find the 5th partial sum, S_5, of the alternating series ∑ (-1)^n / (2n + 1). The detailed solution walks through step-by-step calculations. Suitable for students in advanced high school and early college levels.

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