I know that the series $\sum\limits_{n=1}^{\infty} \dfrac{(-1)^n}{4n-3}$ is convergent by Leibniz's law. However, finding the exact sum of this series can be quite challenging.
I try to evaluate out to $10$ terms is $0.861229$; $20$ terms is $0.863979$. Would you please share me a way to solve this question?
Best wishes,
Take the partial sums $$ \sum_{n=0}^N\frac{(-1)^{n+1}}{4n+1}=-\sum_{n=0}^N\int_0^1 (-x^4)^ndx=-\int_0^1\frac{1-(-x^4)^{N+1}}{1+x^4}dx. $$ Now, $$ \left|\int_0^1\frac{(-x^4)^{N+1}}{1+x^4}dx\right|\leqslant\int_0^1 x^{4N+4}dx=\frac{1}{4N+5}\underset{N\rightarrow +\infty}{\longrightarrow}0 $$ so, letting $N\rightarrow +\infty$, we get $$ \sum_{n=0}^{+\infty}\frac{(-1)^{n+1}}{4n+1}=-\int_0^1\frac{dx}{1+x^4}. $$ To compute the above integral, you can brute-force with partial fraction decomposition, this gives calculations I don't have the courage to write down (though you can watch this video for details), but in the end you get $$ \sum_{n=0}^{+\infty}\frac{(-1)^{n+1}}{4n+1}=-\frac{\pi+2\log(1+\sqrt{2})}{4\sqrt{2}}. $$