For the following Series, decide whether they converge or diverge. \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2 n-1} \)

Determine the convergence of the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2 n-1} \). Determine the convergence or divergence by following steps: - step0: Determine the convergence or divergence using the Alternating Series Test: \(\sum _{n=1}^{+\infty}\left(-1\right)^{n+1}\times \frac{1}{2n-1}\) - step1: Multiply the terms: \(\sum _{n=1}^{+\infty}\left(-\frac{\left(-1\right)^{n}}{2n-1}\right)\) - step2: Find the limit: \(\lim _{n\rightarrow +\infty}\left(\left|-\frac{\left(-1\right)^{n}}{2n-1}\right|\right)\) - step3: Remove the absolute value bars: \(\lim _{n\rightarrow +\infty}\left(\frac{1}{2n-1}\right)\) - step4: Rewrite the expression: \(\frac{1}{\lim _{n\rightarrow +\infty}\left(2n-1\right)}\) - step5: Calculate: \(\frac{1}{+\infty}\) - step6: Calculate: \(0\) - step7: Rewrite the expression: \(-\frac{2n-1}{2n+1}>-1\) - step8: Change the sign: \(\frac{2n-1}{2n+1}<1\) - step9: Calculate: \(\frac{2n-1}{2n+1}-1<0\) - step10: Calculate: \(-\frac{2}{2n+1}<0\) - step11: Change the sign: \(\frac{2}{2n+1}>0\) - step12: Rewrite the expression: \(2n+1>0\) - step13: Move the constant to the right side: \(2n>0-1\) - step14: Remove 0: \(2n>-1\) - step15: Divide both sides: \(\frac{2n}{2}>\frac{-1}{2}\) - step16: Divide the numbers: \(n>-\frac{1}{2}\) - step17: Calculate: \(\frac{1}{2n-1}\) - step18: Calculate: \(\frac{1}{2n+1}\) - step19: The inequality is true: \(\frac{1}{2n-1}>\frac{1}{2n+1}\) - step20: Determine the convergence or divergence: \(\text{Converges}\) La serie \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{2 n-1} \) converge.