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Show that \(\displaystyle\sum_{k=0}^{n-1}(-1)^{n-k-1}\dfrac{(n+k)!}{(k!)^2(n-k-1)!}=n^2\).

Evaluate \(\displaystyle\sum_{r=1}^{\infty}\dfrac{(-1)^r}{r\binom{n+r}{n}}\).

Show that \(\displaystyle \sum_{k=1}^{n-1}(-1)^k\sin^n\left(\frac{k\pi}{n}\right)=\frac{(1+(-1)^n)n}{2^n}\cos\left(\frac{n\pi}{2}\right).\)

(Ovidiu Furdui)

Writing \(a_n=(-1)^{n+1}\cos\frac{1}{n}\) we have that \(a_{2k}\to-1\) so \(a_n\not\to0\) and consequently the series doesn't converge.

We have \(\begin{aligned}n^a(\sqrt{n+1}-2\sqrt{2}+\sqrt{n+1})&=n^{a+1/2}(\sqrt{1+1/n}-2+\sqrt{1-1/n})\\&=n^{a+1/2}(1+\frac{1}{2n}-\frac{1}{8n^2}-2+1-\frac{1}{2n}-\frac{1}{8n^2}+\mathcal O(n^{-3}))\\&=-\frac{1}{4n^{3/2-a}}+\mathcal O(n^{a-5/2})\,,\end{aligned}\) so the series converges fo...

Taking logarithms, it suffices to show that \[-\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k}\right)>\frac{1}{2}\ln 2n\,.\] But \[-\sum_{k=1}^{n}\ln\left(1-\frac{1}{2k}\right)=\sum_{k=1}^{n}\sum_{m=1}^{+\infty}\frac{(2k)^{-m}}{m}>\sum_{k=1}^{n}\sum_{m=1}^{2}\frac{(2k)^{-m}}{m}=\frac{H_{n}}{2}+\frac{1}{4}\sum...