calculate-S-1-1-2-1-3-4-1-5-6-

Question Number 63101 by mathmax by abdo last updated on 29/Jun/19

c a l c u l a t e S = \frac{1}{1 \times 2} + \frac{1}{3 \times 4} + \frac{1}{5 \times 6} + \dots . .

Commented by mathmax by abdo last updated on 29/Jun/19

l e t t r y a n o t h e r w a y w e h a v e \frac{d}{d x} l n (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} w i t h ∣ x ∣< 1 \Rightarrow

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

{l i m}_{x \to 1} l n (1 + x) = l n (2) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots =

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

S = l n (2) .

Answered by mr W last updated on 29/Jun/19

S = \frac{1}{1 \times 2} + \frac{1}{3 \times 4} + \frac{1}{5 \times 6} + \dots . . > 0

S = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{5} - \frac{1}{6}) +

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

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

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

S = 0 ?

Commented by MJS last updated on 29/Jun/19

I seem to remember that {it}^{'} s not allowed

to rearrange in this way

S = \frac{1}{2} + s \Rightarrow S > \frac{1}{2}

S = \frac{7}{12} + s \Rightarrow S > \frac{7}{12}

\dots

Commented by turbo msup by abdo last updated on 29/Jun/19

w e h s v e S = Σ a_{i} a n d a_{i} > 0 h o w

c a n S b e 0 \dots .

Commented by turbo msup by abdo last updated on 29/Jun/19

w e h s v e S = Σ a_{i} a n d a_{i} > 0 h o w

c a n S b e 0 \dots .

Commented by mathmax by abdo last updated on 29/Jun/19

s i r m r w y o u r b e g i n i n g i s c o r r e c t b u t t h e a n s w e r i s n o t 0 l e t c o m p l e t e t h e

w o r k w e h a v e S = {l i m}_{n \to + \infty} (\sum_{k = 0}^{n} \frac{1}{2 k + 1} - \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{k})

w e h a v e \sum_{k = 1}^{n} \frac{1}{k} = H_{n}

\sum_{k = 0}^{n} \frac{1}{2 k + 1} = 1 + \frac{1}{3} + \frac{1}{5} + \dots . + \frac{1}{2 n + 1} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots . + \frac{1}{2 n} + \frac{1}{2 n + 1}

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

\sum_{k = 0}^{n} \frac{1}{2 k + 1} - \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{k} = H_{2 n + 1} - H_{n} = l n (2 n + 1) + γ + o (\frac{1}{n}) - l n (n) - γ + o (\frac{1}{n})

= l n (\frac{2 n + 1}{n}) + o (\frac{1}{n}) \to l n (2) (n \to + \infty) s o

\frac{1}{1 \times 2} + \frac{1}{3 \times 4} + \frac{1}{5 \times 6} + \dots . . = l n (2) . (γ i s t h e c o n s t a n t o f e u l e r)

Commented by mr W last updated on 29/Jun/19

t h a n k s s i r! i l e a r n t s o m e t h i n g n e w .

Commented by mathmax by abdo last updated on 29/Jun/19

y o u a r e a l w a y s w e l c o m e .

Answered by MJS last updated on 29/Jun/19

it seems to be

S = \ln 2

but I cannot properly show it

Answered by naka3546 last updated on 29/Jun/19

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