Question Number 65386 by mathmax by abdo last updated on 29/Jul/19
$${calculate}\:\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{99}{k}−\mathrm{1}} \\ $$
Commented by mathmax by abdo last updated on 31/Jul/19
![let I =∫_1 ^(+∞) (dx/(1+x^(99) )) cha7gement x=(1/t) give I =−∫_0 ^1 (1/(1+(1/t^(99) )))(−(dt/t^2 )) = ∫_0 ^1 (t^(99) /(t^2 (1+t^(99) )))dt =∫_0 ^1 (t^(97) /(1+t^(99) ))dt =∫_0 ^1 t^(97) Σ_(k=0) ^∞ (−1)^k t^(99k) dt = Σ_(k=0) ^∞ (−1)^k ∫_0 ^1 t^(99k+97) dt =Σ_(k=0) ^∞ (−1)^k [(1/(99k+98)) t^(99k+98) ]_0 ^1 =Σ_(k=0) ^∞ (((−1)^k )/(99k+98)) =_(k=p−1) Σ_(p=1) ^∞ (((−1)^(p−1) )/(99(p−1)+98)) =Σ_(p=1) ^∞ (((−1)^(p−1) )/(99p−1)) ⇒ Σ_(k=1) ^∞ (((−1)^(k−1) )/(99k−1)) =∫_1 ^(+∞) (dx/(1+x^(99) )) ∫_1 ^(+∞) (dx/(1+x^(99) )) =∫_1 ^0 (....)dx +∫_0 ^∞ (dx/(1+x^(99) )) =∫_0 ^∞ (dx/(1+x^(99) ))−∫_0 ^1 (dx/(1+x^(99) )) changement x =t^(1/(99)) give ∫_0 ^∞ (dx/(1+x^(99) )) =∫_0 ^∞ (1/(1+t))(1/(99))t^((1/(99))−1) dt =(1/(99))∫_0 ^∞ (t^((1/(99))−1) /(1+t))dt =(1/(99)) (π/(sin((π/(99))))) =(π/(99sin((π/(99))))) ∫_0 ^1 (dx/(1+x^(99) )) =∫_0 ^1 Σ_(n=0) ^∞ (−1)^n x^(99n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(99n) dx =Σ_(n=0) ^∞ (−1)^(n ) [(1/(99n+1)) x^(99n+1) ]_0 ^1 =Σ_(n=0) ^∞ (((−1)^n )/(99n+1))=S =1−(1/(100)) +(1/(199)) −(1/(3.99+1)) +....⇒ Σ_(k=1) ^∞ (((−1)^(k−1) )/(99k−1)) =(π/(99sin((π/(99))))) −S](https://www.tinkutara.com/question/Q65584.png)
$${let}\:{I}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} }\:\:{cha}\mathrm{7}{gement}\:\:{x}=\frac{\mathrm{1}}{{t}}\:{give}\: \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{99}} }}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{\mathrm{99}} }{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{99}} \right)}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{\mathrm{97}} }{\mathrm{1}+{t}^{\mathrm{99}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{97}} \sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} {t}^{\mathrm{99}{k}} \:{dt}\:=\:\sum_{{k}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{\mathrm{99}{k}+\mathrm{97}} \:{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \left[\frac{\mathrm{1}}{\mathrm{99}{k}+\mathrm{98}}\:{t}^{\mathrm{99}{k}+\mathrm{98}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{99}{k}+\mathrm{98}} \\ $$$$=_{{k}={p}−\mathrm{1}} \:\:\:\:\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{99}\left({p}−\mathrm{1}\right)+\mathrm{98}}\:=\sum_{{p}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{99}{p}−\mathrm{1}}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{99}{k}−\mathrm{1}}\:=\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} } \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} }\:=\int_{\mathrm{1}} ^{\mathrm{0}} \left(….\right){dx}\:+\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} }−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} } \\ $$$${changement}\:{x}\:={t}^{\frac{\mathrm{1}}{\mathrm{99}}} \:{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\frac{\mathrm{1}}{\mathrm{99}}{t}^{\frac{\mathrm{1}}{\mathrm{99}}−\mathrm{1}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{99}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{99}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\frac{\mathrm{1}}{\mathrm{99}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{99}}\right)}\:=\frac{\pi}{\mathrm{99}{sin}\left(\frac{\pi}{\mathrm{99}}\right)} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{99}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{99}{n}} {dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{99}{n}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}\:} \left[\frac{\mathrm{1}}{\mathrm{99}{n}+\mathrm{1}}\:{x}^{\mathrm{99}{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{99}{n}+\mathrm{1}}={S} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{100}}\:+\frac{\mathrm{1}}{\mathrm{199}}\:−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{99}+\mathrm{1}}\:+….\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{99}{k}−\mathrm{1}}\:=\frac{\pi}{\mathrm{99}{sin}\left(\frac{\pi}{\mathrm{99}}\right)}\:−{S}\:\: \\ $$