Summing An Interesting Infinite Series

[@yurenchu]

It seems to me the answer at the end is incorrect. The correct answer would be (1/3)*[ ln(2) + *2* * (√3)*arctan(1/√3) ] . Because at 4:30 , the RHS was integrated from x=0 to x=1 , and therefore, the LHS must also be integrated from x=0 to x=1 . EDIT: I just asked Wolfram Alpha, and Wolfram Alpha agrees with me: 1 - 1/4 + 1/7 - 1/10 + ... = (1/9)*( √3 π + 3*ln(2) ) ≈ 0.83565 which means the series is equal to (1/3)*ln(2) + (1/3)*(√3)*(π/3) = = (1/3)*[ ln(2) + (√3)*arctan(√3) ] = (1/3)*[ ln(2) + (√3)*2*arctan(1/√3) ]

[@bobbyheffley4955]

The series obtained by taking the absolute value of the given series diverges. The original series is conditionally convergent.

[@reconquistahinduism346]

Scientific calculator pegs the value at 0.3.....

[@meurdesoifphilippe5405]

Arctan(1/sqrt(3))... you mean pi/6?

[@SyberMath]

Yes

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@RealQinnMalloryu4]

(1)^2 (1)^2/(16)=1 ➖ 1/16 ={0+0 ➖ }/16={1n+1n ➖}/{16n+16n ➖}+{1n+1n ➖}/{7n+7n ➖ }={2n^2/32n^2+2n^2/14n^2}=4n^4/46n^4 ➖ (1)^2/(10)^2={4n^4/46n^4 ➖ 1/100}=3n^4/64n^4+{1n+1n ➖}/{13n+13n ➖}={3n^4/64n^4+2n^2/26n^2 }=5n^6/90n^6 ➖ (1)^2/(16)^2={5n^6/90n^6 ➖ 1/256}=4n^6/166n^6+{1n+1n ➖}/{19n+19n ➖ }={4n^6/166n^6+2n^2/38n^2}=6n^8/204n^8 6n^8/38^166n^8 63^3n^2^3/1^3^3^3^3n^2^3 1^1n^1^1^1/1^1^1^1^3n^1^1^2 /1^3n^1^2 /3n^2 (n ➖ 3n+2).

[@blabberblabbing8935]

How are 1/(1+x^3) or (-x)^(3n) even related to (-1)^n/(1+3n)? 🤔