including series

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$\sum_{a=1}^{n}k^a=\frac{k(k^n-1)}{k-1}~~~~~~~~(k\neq1)$ 
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$\sum_{a=1}^{n}ak^a= \frac{k\{nk^{n+1}-(n + 1)k^n+1\}}{(k - 1)^2}~~~~~~~~(k\neq1)$
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$\sum_{a=1}^{n} a = \frac{n(n+1)}{2} $
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$\sum_{a=1}^{n} a^2 = \frac{n(n+1)(2n+1)}{6} $
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$\sum_{a=1}^{n} a^3 = (\frac{n(n+1)}{2})^2 $
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$\sum_{a=1}^{n} a^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} $
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$\sum_{i=1}^{\infty} \frac{(-1)^{i-1}}{2i-1}=\frac{\pi}{4}$   (Leibniz)
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$\sum_{j=0}^{n} \sin(\alpha+j\beta)=\frac{\sin(\alpha+\frac{n}{2}\beta) \sin\frac{(n+1)\beta}{2} }{\sin \frac{\beta}{2}}$
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$\sum_{j=0}^{n} \cos(\alpha+j\beta)=\frac{\cos(\alpha+\frac{n}{2}\beta) \sin\frac{(n+1)\beta}{2} }{\sin \frac{\beta}{2}}$

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$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

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$\sum_{n=1}^\infty \frac{1}{n^3}=\frac{2\pi^2\log{2}}{7}+\frac{16}{7}\int_{0}^{\frac{\pi}{2}}x\log{(\sin{x})}dx$

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$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$

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$\sum_{n=1}^\infty \frac{1}{n^6}=\frac{\pi^6}{945}$

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$\sum_{n=1}^\infty \frac{1}{n^8}=\frac{\pi^8}{9450}$

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$\sum_{n=1}^\infty \frac{1}{n^10}=\frac{\pi^10}{93555}$

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$\sum_{n=1}^\infty \frac{1}{n^{12}}=\frac{691\pi^{12}}{638512875}$

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$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$

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$\sum_{n=1}^{\infty}\frac{1}{n^3(n^3+1)}=10-\pi^2$

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$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^4}=\frac{\pi^4}{96}$
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$\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}=2-2\log{2}$
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$\sum_{n=1}^{\infty} \frac{1}{n(4n^2-1)}=2\log{2}-1$
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$\sum_{n=0}^{\infty}\frac{1}{\{3(2n-1)\}^3-3(2n-1)}=\frac{\log{3}}{4}-\frac{\log{2}}{3}$

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 $\sum_{n=1}^\infty \frac{1}{n^4(n+1)^4}=-35+\frac{10\pi^2}{3}+\frac{\pi^4}{45}$
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$\sum_{n=0}^{\infty}\frac{(-1)^n(2n)!}{2^{2n}(n!)^2}\sin(2n+1)x=\frac{\displaystyle\sin{(\frac{x}{2})}}{\sqrt{2\cos{x}}}$

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$\sum_{n=1}^{\infty} \frac{1}{n!}=e$
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$\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2n-1}}{(2n-1)!}=\sin{x}$
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$\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n}}{2n!}=\cos{x}$
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$\sum_{n=0}^{\infty} \frac{x^{n}}{n!}=e^{x}$
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$\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2-x^2}=\frac{\pi}{4x}\tan{\frac{{\pi}x}{2}}$
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$\sum_{k=1}^{\infty} \log{(1-\frac{4x^2}{(2k-1)^2\pi^2})}=\log{\cos{x}}$  (-\frac{\pi}{2}
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 $\sum_{k=1}^\infty \arctan{\frac{2x^2}{k^2}}=\frac{\pi}{4}-\arctan{\frac{\tanh{\pi x}}{\tan{\pi x}}}$
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$\sum_{n=1}^{\infty}\frac{n}{e^{2{\pi}n}-1}=\frac{1}{24}-\frac{1}{8\pi}$
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$\sum_{n=1}^{\infty}\frac{n^5}{e^{2{\pi}n}-1}=\frac{1}{504}$

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$\sum_{n=0}^{\infty}(-1)^nx^{6n^2+4n}(1+x^{4n+2})=\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{\displaystyle\prod_{m=0}^{n}(1-x^{2m+1})}~~~~~~~~~~|x|<1$

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$\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}\frac{1}{(m+ni)^8}=\frac{1}{525}(2\int_{0}^{2}\frac{dx}{\sqrt{1-x^4}})^8$

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