prove-that-k-1-1-k-2k-1-2-2ln-2-

Question Number 63574 by Tawa1 last updated on 05/Jul/19

prove that \underset{k = 1}{\sum^{\infty}} \frac{1}{k (2 k + 1)} = 2 - 2 \ln (2)

Commented by mathmax by abdo last updated on 05/Jul/19

t h e H_{n} m e t h o d l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{k (2 k + 1)} \Rightarrow \frac{1}{2} S_{n} = \sum_{k = 1}^{n} \frac{1}{2 k (2 k + 1)}

= \sum_{k = 1}^{n} {\frac{1}{2 k} - \frac{1}{2 k + 1}} = \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{k} - \sum_{k = 1}^{n} \frac{1}{2 k + 1} w e h a v e

\sum_{k = 1}^{n} \frac{1}{k} = H_{n}

\sum_{k = 1}^{n} \frac{1}{2 k + 1} = \frac{1}{3} + \frac{1}{5} + \dots \dots + \frac{1}{2 n + 1} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots . + \frac{1}{2 n} + \frac{1}{2 n + 1}

- 1 - \frac{1}{2} - \frac{1}{4} - \dots . - \frac{1}{2 n} = H_{2 n + 1} - 1 - \frac{1}{2} H_{n} \Rightarrow

\frac{1}{2} S_{n} = \frac{1}{2} H_{n} - H_{2 n + 1} + 1 + \frac{1}{2} H_{n} = H_{n} - H_{2 n + 1} + 1

= l n (n) + γ + o (\frac{1}{n}) - l n (2 n + 1) - γ + o (\frac{1}{n}) + 1

= l n (\frac{n}{2 n + 1}) + o (\frac{1}{n}) + 1 \Rightarrow S_{n} = 2 l n (\frac{n}{2 n + 1}) + 2 \Rightarrow

{l i m}_{n \to + \infty} S_{n} = 2 l n (\frac{1}{2}) + 2 = - 2 l n (2) + 2 . t h e r e s u l t i s p r o v e d .

Commented by Prithwish sen last updated on 05/Jul/19

= Σ [\frac{1}{k} - \frac{2}{2 k + 1}]

= (1 + \frac{1}{2} + \frac{1}{3} + \dots \dots \dots . \infty) - 2 (\frac{1}{3} + \frac{1}{5} + \dots \dots \infty)

= (1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} - \dots \dots . \infty)

= 2 - (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \dots . \infty)

= 2 - \ln (2)

I . {can}^{'} t find out my mistake . Please help anyone

Commented by Tawa1 last updated on 05/Jul/19

God bless you sirs

Commented by mathmax by abdo last updated on 05/Jul/19

m e t h o d o f s e r i e s l e t S = \sum_{n = 1}^{\infty} \frac{1}{n (2 n + 1)} a n d S (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n + 1}}{n (2 n + 1)}

w e h a v e S = S (1) w e h a v e \frac{d S}{d x} (x) = \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n} = w (x^{2}) w i t h w (t) = \sum_{n = 1}^{\infty} \frac{t^{n}}{n}

w e h a v e w^{^{'}} (t) = \sum_{n = 1}^{\infty} t^{n - 1} = \sum_{n = 0}^{\infty} t^{n} = \frac{1}{1 - t} \Rightarrow w (t) = - l n ∣ 1 - t ∣ + c

c = w (0) = 0 \Rightarrow \frac{d S}{d x} (x) = - l n (1 - x^{2}) (w e t a k e - 1 < x < 1) \Rightarrow

S (x) = - \int_{0}^{x} l n (1 - t^{2}) d t + λ (λ = S (0) = 0) \Rightarrow S (x) = - \int_{0}^{x} l n (1 - t^{2}) d t \Rightarrow

S = - \int_{0}^{1} l n (1 - t^{2}) d t = - \int_{0}^{1} l n (1 - t) - \int_{0}^{1} l n (1 + t) d t

\int_{0}^{1} l n (1 - t) d t =_{1 - t = u} \int_{0}^{1} l n (u) d u = {[u l n (u) - u]}_{0}^{1} = - 1

\int_{0}^{1} l n (1 + t) d t =_{1 + t = u} \int_{1}^{2} l n (u) d u = {[u l n (u) - u]}_{1}^{2} = 2 l n (2) - 2 + 1 \Rightarrow

S = 1 - 2 l n (2) + 2 - 1 = 2 - 2 l n (2) .

Commented by Tawa1 last updated on 05/Jul/19

God bless you sir

Commented by Tawa1 last updated on 05/Jul/19

Commented by Tawa1 last updated on 05/Jul/19

Please help sir

Commented by Tawa1 last updated on 05/Jul/19

I appreciate your time sir

Commented by Prithwish sen last updated on 06/Jul/19

What is the answer ?

Commented by Tawa1 last updated on 06/Jul/19

I {don}^{'} t know the answer sir . But you can help sir .

Commented by Prithwish sen last updated on 06/Jul/19

a_{n + 1} = \sqrt{1 + n \sqrt{1 + (n - 1) \sqrt{1 + (n - 2) \sqrt{1 + \dots .}}}}

Now, n - 1 = \sqrt{{(n - 1)}^{2}} = \sqrt{1 + n (n - 2)}

= \sqrt{1 + n \sqrt{{(n - 2)}^{2}}} = \sqrt{1 + n \sqrt{1 + (n - 1) (n - 2)}}

= \sqrt{1 + n \sqrt{1 + (n - 1) \sqrt{1 + (n - 2) \sqrt{1 + . .}}}}

as n \to \infty a_{n + 1} \to (n - 1) \Rightarrow a_{n} \to (n - 2) \Rightarrow \frac{a_{n}}{n} \to (1 - \frac{2}{n}) \to 1

Ω = li \underset{n \to \infty}{m} {[\frac{1}{n} {\frac{1}{(n + \frac{a_{n}}{n})} + \frac{2}{(n + 2 . \frac{a_{n}}{n})} + \dots . . + \frac{n}{(n + n . \frac{a_{n}}{n})}}]}^{n^{\frac{1}{n}} - 1}

= li \underset{n \to \infty}{m} {[\frac{1}{n} {\frac{l}{n + 1} + \frac{2}{n + 2} + \dots \dots \dots . + \frac{n}{n + n}}]}^{n^{\frac{1}{n} - 1}}

= li \underset{n \to \infty}{m} {[\frac{1}{n} {\frac{\frac{1}{n}}{1 + \frac{1}{n}} + \frac{\frac{2}{n}}{1 + \frac{2}{n}} + \dots . . + \frac{\frac{n}{n}}{1 + \frac{n}{n}}}]}^{n^{\frac{1}{n} - 1}}

taking \log

\ln (Ω) = li \underset{n \to \infty}{m} (n^{\frac{1}{n}} - 1) \ln [\lim_{n \to \infty} \frac{1}{n} \underset{r = 1}{\sum^{n}} \frac{\frac{r}{n}}{1 + \frac{r}{n}}]

= li \underset{n \to \infty}{m} (n^{\frac{1}{n}} - 1) \ln [\int_{0}^{1} \frac{xdx}{1 + x}]

= \lim_{n \to \infty} (n^{\frac{1}{n}} - 1) \ln (1 - \ln 2)

Now as n \to \infty \ln Ω \to 0 \Rightarrow Ω \to 1

I {don}^{'} t know whether it is corrector not . Please

give your valuable suggestion .

Commented by Prithwish sen last updated on 06/Jul/19

thanks abdo sir