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Question Number 153518 by Tawa11 last updated on 08/Sep/21

Show whether \underset{n = 1}{\sum^{\infty}} (\frac{x^{2}}{3 + n^{2} x^{2}}) is uniformly convegence for real

value of x .

Answered by puissant last updated on 08/Sep/21

l e t f_{n} (x) = \frac{x^{2}}{3 + n^{2} x^{2}}

f_{n} (\frac{1}{n}) = \frac{{(\frac{1}{n})}^{2}}{3 + n^{2} \times \frac{1}{n^{2}}} = \frac{\frac{1}{n^{2}}}{4} = \frac{1}{4 n^{2}} . .

∣∣ f_{n} ∣ ∣_{\infty} ⩾ \frac{1}{4 n^{2}} \to 0 w h e n n \to \infty

h e n c e \underset{n = 1}{\sum^{\infty}} (\frac{x^{2}}{3 + n^{2} x^{2}}) i s u n i f o r m l y

c o n v e r v e n c e . .

Commented by Tawa11 last updated on 08/Sep/21

God bless you sir . I appreciate .

Answered by mindispower last updated on 08/Sep/21

= \sum_{n ⩾ 1} \frac{1}{(n^{2} + \frac{3}{x^{2}})}, x = \frac{\sqrt{3}}{y}

= \sum_{n ⩾ 1} \frac{1}{(n^{2} + y^{2})}

= \sum_{n ⩾ 0} \frac{1}{(n + 1 + i y) (n + 1 - i y)} = \frac{Ψ (1 + i y) - Ψ (1 - i y)}{2 i y}

Ψ (1 + i y) = Ψ (i y) + \frac{1}{i y}

= - \frac{1}{2 y^{2}} - \frac{Ψ (1 - i y) - Ψ (i y)}{2 i y}

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

\sum_{n ⩾ 1} \frac{x^{2}}{n^{2} x^{2} + 3} = - \frac{3}{2} x^{2} + \frac{π}{2} \frac{\sqrt{3}}{x} t h (\frac{π \sqrt{3}}{x})

Commented by Tawa11 last updated on 08/Sep/21

God bless you sir . I appreciate .

Commented by Tawa11 last updated on 08/Sep/21

What is th sir .

Commented by mindispower last updated on 08/Sep/21

t h (x) = \frac{s h (x)}{c h (x)} = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}

Commented by Tawa11 last updated on 08/Sep/21

Ohh . Thanks sir

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