It doesn't telescopes

Calculus Level 3

1 1 × 2 + 1 3 × 4 + 1 5 × 6 + ⋯ \large \frac{1}{1\times 2} + \frac{1}{3\times 4} +\frac{1}{5\times 6} + \cdots Evaluate the infinite sum above correct to 3 decimal places.


The answer is 0.693.

Vilakshan Gupta by Vilakshan Gupta , 19, India

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3 solutions

S = 1 1 × 2 + 1 3 × 4 + 1 5 × 6 + ⋯ = 1 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ⋯ Since ln ⁡ ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + x 5 5 − x 6 6 + ⋯ = ln ⁡ 2 ≈ 0.693 \begin{aligned} S & = {\color{#3D99F6}\frac 1{1 \times 2}} +{\color{#D61F06}\frac 1{3\times 4}} + {\color{#3D99F6}\frac 1{5\times 6}} + \cdots \\ & = {\color{#3D99F6}\frac 11 - \frac 12} + {\color{#D61F06}\frac 13 - \frac 14} + {\color{#3D99F6}\frac 15 - \frac 16} + \cdots & \small \color{#3D99F6} \text{Since } \ln(1+x) = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \frac {x^5}5 - \frac {x^6}6 + \cdots \\ & = \ln 2 \approx \boxed{0.693} \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you tell me where I went wrong? I think that this sequence can be presented in this format:

∑ n = 1 ∞ 1 n ( n + 1 ) \sum_{n=1}^{\infty} \frac{1}{n(n+1)}

And by trying out no.s instead of infinity from 10 to 1000, I noticed that it converges to 1. Am I mistaken?

James Bacon - 2 years, 11 months ago

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It should be ∑ n = 1 ∞ 1 ( 2 n − 1 ) 2 n \displaystyle \sum_{n=1}^\infty \frac 1{(2n-1)2n} .

Chew-Seong Cheong - 2 years, 11 months ago

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Thanks. How did you find this?

James Bacon - 2 years, 11 months ago

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@James Bacon – When n = 1 n=1 , ( 2 n − 1 ) 2 n = 1 × 2 (2n-1)2n=1\times 2 . When n = 2 n=2 , ( 2 n − 1 ) 2 n = 3 × 4 (2n-1)2n=3\times 4 . When n = 3 n=3 , ( 2 n − 1 ) 2 n = 5 × 6 (2n-1)2n=5\times 6 ....

Chew-Seong Cheong - 2 years, 11 months ago

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@Chew-Seong Cheong – Thanks alot

James Bacon - 2 years, 11 months ago

https://brilliant.org/discussions/thread/help-why-isnt-this-latex-working/

Syed Hamza Khalid - 2 years, 11 months ago
X X
Jul 6, 2018

1 1 × 2 + 1 3 × 4 + 1 5 × 6 + ⋯ = ( 1 − 1 2 ) + ( 1 3 − 1 4 ) + ( 1 5 − 1 6 ) + ⋯ = ln ⁡ 2 \frac{1}{1\times 2} + \frac{1}{3\times 4} +\frac{1}{5\times 6} + \cdots=(1-\frac12)+(\frac13-\frac14)+(\frac15-\frac16)+\cdots=\ln2

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you conclude that it would be ln ⁡ 2 \ln 2 ?

Vilakshan Gupta - 2 years, 11 months ago

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See this

X X - 2 years, 11 months ago
Aaghaz Mahajan
Jul 6, 2018

Manipulate to simply get the Taylor series of ln(2)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I know that if we substitute 1 in the expansion of ln ⁡ ( 1 + x ) \ln(1+x) ,we get the same series, but how can we prove it to be ln ⁡ 2 \ln 2 from the other side...I mean going from L.H.S to R.H.S....

Vilakshan Gupta - 2 years, 11 months ago

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@Vilakshan Gupta Ummm.......I don't get it..tu kya puch rhaa hai??? I mean, see Chew Seong Sir's solution......is that what u r asking for??? Kyonki, we already know that the value of the series is ln(2).......going from LHS to RHS doesn't make sense.....

Aaghaz Mahajan - 2 years, 11 months ago

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