I want to evaluate $\,\displaystyle I_{n}=\int_{0}^{\frac{\pi}{4}} \tan^{2n}(x)\,\mathrm dx$.
I proved that $\,I_{n}+I_{n-1}=\dfrac{1}{2n-1}\,,\,$ where $I_{0}=\dfrac{\pi}{4}$.
From that I found that (it's easy to prove by induction) $$I_{n}=(-1)^{n} \frac{\pi}{4}+\sum_{k=1}^{n} (-1)^{k+1} \frac{1}{2(n-k)+1}$$ My question is, is there any way to telescope that sum ? Any help will be greatly appreciated.
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2$\begingroup$ Those are the remainders of this series I might be remembering wrong, but I think the partial sums of that series cannot be written as a fixed sum of hypergeometric terms. You could express it in terms of other special functions Whether that is any more useful than the original expression or not depends on the final goal. $\endgroup$– plopCommented Jan 4, 2023 at 21:38
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4$\begingroup$ Your formula for $I_n - I_{n-1}$ is wrong. $\endgroup$– Robert IsraelCommented Jan 4, 2023 at 21:45
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3$\begingroup$ $I_{n}-I_{n-1}=\dfrac{1}{2n-1}$ should be $I_{n}+I_{n-1}=\dfrac{1}{2n-1}$. $\endgroup$– xpaulCommented Jan 4, 2023 at 21:49
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2$\begingroup$ Bakugany, you should write $\;I_n+I_{n-1}=\dfrac1{2n-1}\;$ instead of $\;I_n-I_{n-1}=\dfrac1{2n-1}\;$ which is wrong. $\endgroup$– AngeloCommented Jan 4, 2023 at 21:51
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1$\begingroup$ I think that it would be better to write your (correct) formula for $I_n$ in the following way :$$I_n=(-1)^n\left[\dfrac{\pi}4-\sum_{k=1}^n(-1)^{k+1}\dfrac1{2k-1}\right]$$ So we can get that $$I_n=(-1)^n\sum_{k=n+1}^\infty(-1)^{k+1}\dfrac1{2k-1}$$ that is $$I_n=\sum_{k=n+1}^\infty(-1)^{k-n+1}\dfrac1{2k-1}$$ or also $$I_n=\sum_{h=1}^\infty(-1)^{h+1}\dfrac1{2(n+h)-1}$$ $\endgroup$– AngeloCommented Jan 5, 2023 at 1:01
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