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Question Number 152546 by liberty last updated on 29/Aug/21

Answered by qaz last updated on 29/Aug/21

1−(1/(5∙3^2 ))−(1/(7∙3^3 ))+(1/(11∙3^5 ))+(1/(13∙3^6 ))−(1/(17∙3^8 ))−(1/(19∙3^9 ))+... =(1−(1/(7∙3^3 ))+(1/(13∙3^6 ))−(1/(19∙3^9 ))+...+(((−1)^(n−1) )/([1+6(n−1)]∙3^(3(n−1)) ))+...)−((1/(5∙3^2 ))−(1/(11∙3^5 ))+(1/(17∙3^8 ))−...+(((−1)^(n−1) )/([5+6(n−1)]∙3^(2+3(n−1)) ))+...) =Σ_(n=0) ^∞ ((((−1)^n )/((1+6n)∙3^(3n) ))−(((−1)^n )/((5+6n)∙3^(2+3n) ))) =Σ_(n=0) ^∞ (((−1)^n )/(27^n ))((1/(1+6n))−(1/(45+54n))) =Σ_(n=0) ^∞ (((−1)^n )/(27^n ))∫_0 ^1 (x^(6n) −x^(54n+44) )dx =∫_0 ^1 (1/(1+(x^6 /(27))))−(x^(44) /(1+(x^(54) /(27))))dx =...

1−15⋅32−17⋅33+111⋅35+113⋅36−117⋅38−119⋅39+...=(1−17⋅33+113⋅36−119⋅39+...+(−1)n−1[1+6(n−1)]⋅33(n−1)+...)−(15⋅32−111⋅35+117⋅38−...+(−1)n−1[5+6(n−1)]⋅32+3(n−1)+...)=∑∞n=0((−1)n(1+6n)⋅33n−(−1)n(5+6n)⋅32+3n)=∑∞n=0(−1)n27n(11+6n−145+54n)=∑∞n=0(−1)n27n∫01(x6n−x54n+44)dx=∫0111+x627−x441+x5427dx=...

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