The problem is asking to find the sum of the following series:

$\log\left(1 - \frac{1}{2^2}\right) + \log\left(1 - \frac{1}{3^2}\right) + \dots + \log\left(1 - \frac{1}{(n+1)^2}\right) + \dots$

Step-by-Step Solution

1. Express Each Term Individually: Each term in the series has the form: $\log\left(1 - \frac{1}{k^2}\right)$ where $k$ is the index of summation starting from $k = 2$.

2. Simplify the Logarithmic Expression: Notice that $1 - \frac{1}{k^2}$ can be rewritten using the difference of squares: $1 - \frac{1}{k^2} = \frac{(k-1)(k+1)}{k^2}$ So, we have: $\log\left(1 - \frac{1}{k^2}\right) = \log\left(\frac{(k-1)(k+1)}{k^2}\right)$

3. Separate the Logarithmic Terms: Using the properties of logarithms, this can be expanded as: $\log\left(\frac{(k-1)(k+1)}{k^2}\right) = \log(k-1) + \log(k+1) - 2\log(k)$

4. Rewrite the Sum Using Telescoping: Now the sum becomes: $\sum_{k=2}^{n} \left( \log(k-1) + \log(k+1) - 2\log(k) \right)$ When expanded, this is a telescoping series, where many terms cancel out.

5. Simplify the Telescoping Series: After expanding and cancelling terms, we are left with only the boundary terms. Specifically, in the telescoping pattern, all intermediate terms will cancel, leaving terms at the beginning and end.

6. Determine the Final Result: The final result after telescoping should yield a simple expression involving the remaining terms.

Conclusion

This setup allows us to find the sum, and the result will typically depend on the value of $n$, or if it tends towards an infinite sum as $n \to \infty$.

Would you like to go further in simplifying this with specific values or limits?

Related Questions

1. How can we apply telescoping techniques in other series?
2. What happens if we use a different base for the logarithm in this series?
3. How would the series change if the denominator were $k^3$ instead of $k^2$?
4. Can we approximate the result for large values of $n$?
5. How does this technique apply to integrals involving logarithmic functions?

Tip

For telescoping series, always look for patterns where terms cancel, simplifying the sum significantly.