An interesting Infinite sum

Calculus Level 3

Evaluate

ln ⁡ ( 2 ) + ∑ k = 0 ∞ ( − 1 ) k ( 8 k 2 − 6 k + 1 ) + 2 k + 1 ( 4 k 2 − 1 ) ( 4 k − 1 ) \ln(\sqrt{2}) + \sum_{k=0}^\infty \frac{(-1)^{k}(8k^{2}-6k+1)+2k+1}{(4k^{2}-1)(4k-1)}


The answer is 2.57.

Cristiano Sansó by Cristiano Sansó , 16, Italy

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1 solution

Cristiano Sansó
Apr 7, 2020

First of all, we can simplify something

ln ⁡ ( 2 ) + ∑ k = 0 ∞ ( − 1 ) k ( 8 k 2 − 6 k + 1 ) + 2 k + 1 ( 4 k 2 − 1 ) ( 4 k − 1 ) = \ln(\sqrt{2}) + \sum_{k=0}^\infty \frac{(-1)^{k}(8k^{2}-6k+1)+2k+1}{(4k^{2}-1)(4k-1)}=

= ln ⁡ ( 2 ) 2 + ∑ k = 0 ∞ 2 k + 1 + ( − 1 ) k ( 4 k − 1 ) ( 2 k − 1 ) ( 2 k + 1 ) ( 2 k − 1 ) ( 4 k − 1 ) = =\frac{\ln(2)}{2} + \sum_{k=0}^\infty \frac{2k+1+(-1)^{k}(4k-1)(2k-1)}{(2k+1)(2k-1)(4k-1)}=

= ln ⁡ ( 2 ) 2 + ∑ k = 0 ∞ 2 k + 1 ( 2 k + 1 ) ( 2 k − 1 ) ( 4 k − 1 ) + ∑ k = 0 ∞ ( − 1 ) k ( 2 k − 1 ) ( 4 k − 1 ) ( 2 k + 1 ) ( 2 k − 1 ) ( 4 k − 1 ) = =\frac{\ln(2)}{2} + \sum_{k=0}^\infty \frac{2k+1}{(2k+1)(2k-1)(4k-1)}+\sum_{k=0}^\infty \frac{(-1)^{k}(2k-1)(4k-1)}{(2k+1)(2k-1)(4k-1)}=

= ln ⁡ ( 2 ) 2 + ∑ k = 0 ∞ 1 ( 2 k − 1 ) ( 4 k − 1 ) + ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) = =\frac{\ln(2)}{2} + \sum_{k=0}^\infty \frac{1}{(2k-1)(4k-1)}+\sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)}=

now, if you remember these two infinite sums

∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) = π 4 \sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)}=\frac{\pi}{4}

8 ( ∑ k = 1 ∞ 1 ( 4 k − 2 ) ( 4 k − 1 ) ) + 2 ln ⁡ ( 2 ) = π 8(\sum_{k=1}^\infty \frac{1}{(4k-2)(4k-1)})+2\ln(2)=\pi

you can try to do something with the sum above

⇒ ln ⁡ ( 2 ) 2 + ∑ k = 0 ∞ 1 ( 2 k − 1 ) ( 4 k − 1 ) + ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) = \Rightarrow \frac{\ln(2)}{2} + \sum_{k=0}^\infty \frac{1}{(2k-1)(4k-1)}+\sum_{k=0}^\infty \frac{(-1)^{k}}{(2k+1)}=

= ln ⁡ ( 2 ) 2 + 2 ∑ k = 0 ∞ 1 ( 4 k − 2 ) ( 4 k − 1 ) + π 4 = =\frac{\ln(2)}{2} + 2\sum_{k=0}^\infty \frac{1}{(4k-2)(4k-1)}+\frac{\pi}{4}=

= ln ⁡ ( 2 ) 2 + 2 ∑ k = 1 ∞ 1 ( 4 k − 2 ) ( 4 k − 1 ) + 2 ∑ k = 0 0 1 ( 4 k − 2 ) ( 4 k − 1 ) + π 4 = =\frac{\ln(2)}{2} + 2\sum_{k=1}^\infty \frac{1}{(4k-2)(4k-1)} + 2\sum_{k=0}^0 \frac{1}{(4k-2)(4k-1)} + \frac{\pi}{4}=

= ln ⁡ ( 2 ) 2 + 2 ∑ k = 1 ∞ 1 ( 4 k − 2 ) ( 4 k − 1 ) + 2 1 ( − 2 ) ( − 1 ) + π 4 = =\frac{\ln(2)}{2} + 2\sum_{k=1}^\infty \frac{1}{(4k-2)(4k-1)} + 2\frac{1}{(-2)(-1)} + \frac{\pi}{4}=

= ln ⁡ ( 2 ) 2 + π − 2 ln ⁡ ( 2 ) 4 + 1 + π 4 = =\frac{\ln(2)}{2} + \frac{\pi - 2\ln(2)}{4} + 1 + \frac{\pi}{4}=

= π 2 + 1 = =\frac{\pi}{2} + 1=

= 2.570796326794... =\boxed{2.570796326794...}

0 Helpful 0 Interesting 3 Brilliant 0 Confused

@Cristiano Sansó , it is not necessary to mention "(answer in decimal notation with 3 significant digit)" because it is mentioned below the answer box. No one will answer in fractions, binary or hexadecimal format unless they are asked to. But you must include ∑ k = 0 ∞ \displaystyle \sum_{\red{k=0}}^\infty .

Chew-Seong Cheong - 1 year, 2 months ago

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