# $\text{Original \& Alternating}$

\begin{aligned} 1 + \dfrac12 + \dfrac13 + \dfrac14 + \dfrac15 + \dfrac16 + \dfrac17 + \dfrac18 + \dots &= \infty \\ 1 - \dfrac12 + \dfrac13 - \dfrac14 + \dfrac15 - \dfrac16 + \dfrac17 - \dfrac18 + \dots &= \ln2 \\ \\ 1 + \dfrac13 + \dfrac15 + \dfrac17 + \dfrac19 + \dfrac{1}{11} + \dfrac1{13} + \dfrac1{15} + \dots &= \infty \\ 1 - \dfrac13 + \dfrac15 - \dfrac17 + \dfrac19 - \dfrac{1}{11} + \dfrac1{13} - \dfrac1{15} + \dots &= \dfrac{\pi}{4} \\ \\ 1 + \dfrac12 + \dfrac14 + \dfrac18 + \dfrac1{16} + \dfrac{1}{32} + \dfrac1{64} + \dfrac1{128} + \dots &= 2 \\ 1 - \dfrac12 + \dfrac14 - \dfrac18 + \dfrac1{16} - \dfrac{1}{32} + \dfrac1{64} - \dfrac1{128} + \dots &= \dfrac23 \\ \\ 1 + \frac1{2^2} + \dfrac1{3^2} + \dfrac1{4^2} + \dfrac1{5^2} + \dfrac1{6^2} + \dfrac1{7^2} + \dfrac1{8^2} + \dots &= \dfrac{\pi^2}{6} \\ 1 - \frac1{2^2} + \dfrac1{3^2} - \dfrac1{4^2} + \dfrac1{5^2} - \dfrac1{6^2} + \dfrac1{7^2} - \dfrac1{8^2} + \dots &= \dfrac{\pi^2}{12} \\ \\ 1 + \frac1{2!} + \dfrac1{3!} + \dfrac1{4!} + \dfrac1{5!} + \dfrac1{6!} + \dfrac1{7!} + \dfrac1{8!} + \dots &= e - 1 \\ 1 - \frac1{2!} + \dfrac1{3!} - \dfrac1{4!} + \dfrac1{5!} - \dfrac1{6!} + \dfrac1{7!} - \dfrac1{8!} + \dots &= 1 - \dfrac1e \\ \\ 1 + \frac1{3!} + \dfrac1{5!} + \dfrac1{7!} + \dfrac1{9!} + \dfrac1{11!} + \dfrac1{13!} + \dfrac1{15!} + \dots &= \sinh1 \\ 1 - \frac1{3!} + \dfrac1{5!} - \dfrac1{7!} + \dfrac1{9!} - \dfrac1{11!} + \dfrac1{13!} - \dfrac1{15!} + \dots &= \sin1 \\ \\ 1 + \frac1{2!} + \dfrac1{4!} + \dfrac1{6!} + \dfrac1{8!} + \dfrac1{10!} + \dfrac1{12!} + \dfrac1{14!} + \dots &= \cosh1 \\ 1 - \frac1{2!} + \dfrac1{4!} - \dfrac1{6!} + \dfrac1{8!} - \dfrac1{10!} + \dfrac1{12!} - \dfrac1{14!} + \dots &= \cos1 \\ \\ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots &= \dfrac{-1}{12} \\ 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots &= \dfrac{1}{4} \\ \end{aligned}

$\text{Source: blackpen}\textcolor{#D61F06}{redpen}$

1 month, 3 weeks ago

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If you sourced BlackpenRedPen, he mentions that the second-last series has value $\dfrac{-1}{8}$ and not $\dfrac{-1}{12}$, although there can be many more values according to algebraic manipulations. Afterall $\infty= \infty+ 1$

- 1 month, 2 weeks ago

I used Ramanujan's summation.

- 1 month, 2 weeks ago

- 1 month, 3 weeks ago

Glad you liked it @Yajat Shamji

- 1 month, 3 weeks ago