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Question Number 145270 by mnjuly1970 last updated on 03/Jul/21

$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{Arccot}(3+\frac{n(n+1)}{3})=?$$$$.....$$

Answered by Olaf_Thorendsen last updated on 03/Jul/21

$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{Arccot}(3+\frac{n(n+1)}{3})$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{Arctan}\left(\frac{1}{3+\frac{n(n+1)}{3}}\right)$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{Arctan}\left(\frac{\frac{1}{3}}{1+\frac{n(n+1)}{9}}\right)$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\mathrm{Arctan}\left(\frac{\frac{n+1}{3}-\frac{n}{3}}{1+\frac{n}{3}\times \frac{(n+1)}{3}}\right)$$$$\mathrm{S}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\mathrm{Arctan}\left(\frac{n+1}{3}\right)-\mathrm{Arctan}\left(\frac{n}{3}\right))$$$$\mathrm{S}=\frac{\pi }{2}-\mathrm{Arctan}\left(\frac{1}{3}\right)$$$$\mathrm{S}=\frac{\pi }{2}-(\frac{\pi }{2}-\mathrm{Arctan}\left(3\right))$$$$\mathrm{S}=\mathrm{Arctan}\left(3\right)$$

Commented by mnjuly1970 last updated on 03/Jul/21

$$verynicemrolaf...$$

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