let-give-u-n-k-0-1-k-1-2-2-k-find-lim-n-u-n-

Question Number 29505 by abdo imad last updated on 09/Feb/18

l e t g i v e u_{n} = \sum_{k = 0}^{\infty} \frac{1}{{(k + 1)}^{2} 2^{k}} f i n d {l i m}_{n \to \infty} u_{n} .

Commented by abdo imad last updated on 11/Feb/18

u_{n} = \sum_{k = 0}^{n} \frac{1}{{(k + 1)}^{2} 2^{k}} .

Commented by abdo imad last updated on 13/Feb/18

w e h a v e u_{n} = \sum_{k = 1}^{n + 1} \frac{1}{k^{2} 2^{k - 1}} = 2 \sum_{k = 1}^{n + 1} \frac{1}{k^{2}} {(\frac{1}{2})}^{k} b u t t h s e r i e

\sum_{k = 1}^{\infty} \frac{1}{k^{2}} {(\frac{1}{2})}^{k} i s c o n v r g e n t e l e t p u t w (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}}

f o r ∣ x ∣< 1 w e h a v e w^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} \Rightarrow {x w}^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

\Rightarrow w^{^{'}} (x) + {x w}^{^{″}} (x) = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} s o w i s

s o l u t i o n f o r t h e d . e y^{^{'}} + {x y}^{^{″}} = \frac{1}{1 - x} f o r y^{^{'}} = z \Rightarrow

z + {x z}^{^{'}} = \frac{1}{1 - x} h . e \Rightarrow z + {x z}^{^{'}} = 0 \Rightarrow {x z}^{^{'}} = - z \Rightarrow \frac{z^{^{'}}}{z} = \frac{- 1}{x}

\Rightarrow l n ∣ z ∣= - l n ∣ x ∣ + k \Rightarrow z = \frac{λ}{x} f o r 0 < x < 1 m v c \Rightarrow

z^{^{'}} = \frac{λ^{^{'}}}{x} - \frac{λ}{x^{2}} \Rightarrow \frac{λ}{x} + λ^{^{'}} - \frac{λ}{x} = \frac{1}{1 - x} \Rightarrow λ^{^{'}} = \frac{1}{1 - x} \Rightarrow

λ (x) = - l n (1 - x) + k a n d k = λ (0) = 0

z (x) = - \frac{l n (1 - x)}{x} = y^{^{'}} \Rightarrow y (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t + c

c = y (0) = 0 \Rightarrow w (x) = - \int_{0}^{x} \frac{l n (1 - t)}{t} d t a n d

S = 2 w (\frac{1}{2}) = - 2 \int_{0}^{\frac{1}{2}} \frac{l n (1 - t)}{t} d t \dots . . b e c o n t i n u e d \dots .

Commented by abdo imad last updated on 14/Feb/18

w e h a v e {l i m}_{n \to \infty} u_{n} = 2 \sum_{n = 1}^{\infty} \frac{1}{n^{2}} {(\frac{1}{2})}^{n} l e t p u t

S (x) = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow \int_{0}^{x} S (t) d t = \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

= - l n (1 - x) f o r 0 < x < 1 (λ = 0) \Rightarrow

x \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} = - l n (1 - x) \Rightarrow \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} = - \frac{l n (1 - x)}{x} \Rightarrow

\int_{0}^{x} (\sum_{n = 1}^{\infty} \frac{t^{n - 1}}{n}) d t = - \int_{0}^{x} \frac{l n (1 - t)}{t} + λ \Rightarrow

\sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}} = - \int_{0}^{x} \frac{l n (1 - t)}{t} (λ = 0) a n d

2 \sum_{n = 1}^{\infty} \frac{1}{n^{2}} {(\frac{1}{2})}^{n} = - \int_{0}^{\frac{1}{2}} \frac{l n (1 - t)}{t} d t c h . t = c o s θ

\int_{0}^{\frac{1}{2}} (\dots) d t = \int_{\frac{π}{2}}^{\frac{π}{3}} \frac{l n (1 - c o s θ)}{c o s θ} s i n θ d θ

= \int_{\frac{π}{2}}^{\frac{π}{3}} t a n θ l n (1 - c o s θ) d θ \dots . b e c o n t i n u e d \dots