n-1-1-n-2-1-n-2-1-n-2-1-

Question Number 121326 by Dwaipayan Shikari last updated on 06/Nov/20

\frac{\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1}}{\underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} - 1}}

Commented by Dwaipayan Shikari last updated on 06/Nov/20

\underset{n = 2}{\sum^{\infty}} \frac{1}{n^{2} - 1} = \frac{1}{2} (\underset{n = 2}{\sum^{\infty}} \frac{1}{n - 1} - \underset{n = 2}{\sum^{\infty}} \frac{1}{n + 1}) = \frac{1}{2} (1 + \frac{1}{2} + \frac{1}{3} + \dots - \frac{1}{3} - \frac{1}{4} - \dots)

= \frac{3}{4}

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1} = \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{1}{n - i} - \frac{1}{n + i}

= \frac{1}{2 i} \underset{n = 2}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n - i - 1}) - \frac{1}{2 i} \underset{n = 2}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + i - 1})

= \frac{1}{2 i} (ψ (i) - ψ (1 - i))

= \frac{1}{2 i} (- π c o t (i π)) + i) = \frac{1}{2 i} (- π c o t (i π) + 1)) = \frac{π}{2} c o t h π + \frac{1}{2}

[Γ (s) Γ (1 - s) = \frac{π}{s i n π s}]

[l o g (Γ (s) + l o g (Γ (1 - s)) = l o g π - l o g (s i n π s)]

[\frac{Γ^{'} (s)}{Γ (s)} - \frac{Γ^{'} (1 - s)}{Γ (1 - s)} = - \frac{π c o s s π s}{s i n π s}]

[ψ (1 - s) - ψ (s) = π c o t π s]

Commented by Dwaipayan Shikari last updated on 06/Nov/20

I s i t r i g h t ?

Commented by mathmax by abdo last updated on 06/Nov/20

perhaps your answer is correct . .!

Answered by mathmax by abdo last updated on 06/Nov/20

let f (x) = e^{- ∣ x ∣} function 2 π periodic even we have

f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos (nx) with a_{n} = \frac{2}{T} \int_{T} f (x) \cos (nx) dx

= \frac{2}{2 π} \int_{- π}^{π} e^{- ∣ x ∣} \cos (nx) dx = \frac{2}{π} \int_{0}^{π} e^{- x} \cos (nx) dx

\Rightarrow \frac{π}{2} a_{n} = Re (\int_{0}^{π} e^{- x + inx} dx) and \int_{0}^{π} e^{(- 1 + in) x} dx

= {[\frac{1}{- 1 + in} e^{(- 1 + in) x}]}_{0}^{π} = \frac{- 1}{1 - in} {e^{(- 1 + in) π} - 1}

= - \frac{1 + in}{1 + n^{2}} {{(- 1)}^{n} e^{- π} - 1} = \frac{1 - {(- 1)}^{n} e^{- π}}{1 + n^{2}} + i (\dots) \Rightarrow

a_{n} = \frac{2}{π} \times \frac{1 - {(- 1)}^{n} e^{- π}}{1 + n^{2}} \Rightarrow

a_{0} = \frac{2}{π} (1 - e^{- π}) \Rightarrow \frac{a_{0}}{2} = \frac{1 - e^{- π}}{π} \Rightarrow

e^{- ∣ x ∣} = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{1 - {(- 1)}^{n} e^{- π}}{1 + n^{2}} \cos (nx) \Rightarrow

x = 0 \Rightarrow 1 = \frac{1 - e^{- π}}{π} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{1}{1 + n^{2}} - \frac{{2 e}^{- π}}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1}

\Rightarrow 1 = \frac{1 - e^{- π}}{π} + \frac{2}{π} x - \frac{{2 e}^{- π}}{π} y

x = π \Rightarrow e^{- π} = \frac{1 - e^{- π}}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{1 - {(- 1)}^{n} e^{- π}}{n^{2} + 1} \Rightarrow

e^{- π} = \frac{1 - e^{- π}}{2} + \frac{2}{π} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2} + 1} - \frac{{2 e}^{- π}}{π} \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1}

= \frac{1 - e^{- π}}{2} + \frac{2}{π} y - \frac{{2 e}^{- π}}{π} x we get the system

{\begin{cases} \frac{2}{π} x - \frac{{2 e}^{- π}}{π} y = 1 - \frac{1 - e^{- π}}{π} \Rightarrow \\ - \frac{{2 e}^{- π}}{π} x + \frac{2}{π} y = e^{- π} - \frac{1 - e^{- π}}{2} \end{cases}

{\begin{cases} 2 x - {2 e}^{- π} y = π - 1 + e^{- π} \\ - {2 e}^{- π} x + 2 y = π e^{- π} - \frac{π - π e^{- π}}{2} \end{cases}

Δ_{s} = 4 - {4 e}^{- 2 π} \Rightarrow x = \frac{Δ_{x}}{Δ}

Δ_{x} = | \begin{matrix} π - 1 + e^{- π} - {2 e}^{- π} \\ π e^{- π} - \frac{π - π e^{- π}}{2} 2 \end{matrix} | = \dots .

x = \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} \dots . be continued \dots