mathematical-analysis-prove-that-n-1-3-n-1-4-n-n-1-pi-m-n-july-1970-

Question Number 114099 by mnjuly1970 last updated on 17/Sep/20

\dots . m a t h e m a t i c a l a n a l y s i s \dots .

p r o v e t h a t ::

\underset{n = 1}{\sum^{\infty}} (\frac{3^{n} - 1}{4^{n}}) ζ (n + 1) = π

You can't use 'macro parameter character #' in math mode

Answered by maths mind last updated on 17/Sep/20

= \sum_{n ⩾ 1} . \sum_{m ⩾ 1} (\frac{3^{n} - 1}{4^{n}}) . \frac{1}{m^{n + 1}}

= \sum_{m ⩾ 1} \frac{1}{m} {\sum_{n ⩾ 1} {(\frac{3}{4 . m})}^{n} - \sum_{n ⩾ 1} {(\frac{1}{4 m})}^{n}}

= \sum_{m ⩾ 1} \frac{1}{m} {\frac{3}{4 m} . \frac{1}{1 - \frac{3}{4 m}} - \frac{1}{4 m} . \frac{1}{1 - \frac{1}{4 m}}}

= \sum_{m ⩾ 1} \frac{1}{m} {\frac{3}{4 m - 3}} - \sum_{m ⩾ 1} \frac{1}{m (4 m - 1)}

= \sum_{m ⩾ 0} \frac{3}{(1 + m) (4 m + 1)} - \sum_{m ⩾ 0} \frac{1}{(m + 1) (4 m + 3)}

= \sum_{m ⩾ 0} \frac{1 - \frac{1}{4}}{(m + 1) (m + \frac{1}{4})} - \sum_{m ⩾ 0} \frac{1 - \frac{3}{4}}{(m + 1) (m + \frac{3}{4})}

= Ψ (1) - Ψ (\frac{1}{4}) - {Ψ (1) - Ψ (\frac{3}{4})}

= Ψ (\frac{3}{4}) - Ψ (\frac{1}{4}) = Ψ (1 - \frac{1}{4}) - Ψ (\frac{1}{4}) = π c o t (\frac{π}{4}) = π

Commented by mnjuly1970 last updated on 18/Sep/20

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