The problem asks to compute the 5th partial sum, $s_5$, of the given series:

$\sum_{n=1}^{\infty} \frac{(-1)^n}{2n+1}$

Specifically, the 5th partial sum is given by:

$s_5 = \sum_{n=1}^{5} \frac{(-1)^n}{2n+1}$

To compute this, we need to calculate the terms for $n = 1$ to $n = 5$ and sum them.

Let's break it down:

1. For $n = 1$: $\frac{(-1)^1}{2(1) + 1} = \frac{-1}{3}$

2. For $n = 2$: $\frac{(-1)^2}{2(2) + 1} = \frac{1}{5}$

3. For $n = 3$: $\frac{(-1)^3}{2(3) + 1} = \frac{-1}{7}$

4. For $n = 4$: $\frac{(-1)^4}{2(4) + 1} = \frac{1}{9}$

5. For $n = 5$: $\frac{(-1)^5}{2(5) + 1} = \frac{-1}{11}$

Now, summing these up:

$s_5 = \left( \frac{-1}{3} \right) + \left( \frac{1}{5} \right) + \left( \frac{-1}{7} \right) + \left( \frac{1}{9} \right) + \left( \frac{-1}{11} \right)$

Let's compute the numerical result.The 5th partial sum, $s_5$, of the series is approximately:

$s_5 \approx -0.256$

Would you like more details or have any questions about this result?

Here are 5 related questions that expand on this topic:

1. What is the general behavior of this alternating series as $n \to \infty$?
2. How does the 5th partial sum compare to the sum of the infinite series?
3. Can this series be related to any well-known functions or integrals?
4. How would the result change if we found the 10th partial sum instead of the 5th?
5. What tests can we use to determine if this series converges?

Tip: Alternating series often converge more quickly than non-alternating ones, especially if the terms decrease in magnitude.