let-I-n-0-n-1-x-2x-1-2-dx-1-calculate-I-n-interms-of-n-2-find-lim-n-I-n-

Question Number 37896 by abdo mathsup 649 cc last updated on 19/Jun/18

$${let}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}_{{n}} \:\:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} \\ $$

Commented by prof Abdo imad last updated on 19/Jun/18

1) I_n = Σ_(k=0) ^(n−1) ∫_k ^(k+1) (((−1)^k )/((2x+1)^2 ))dx =Σ_(k=0) ^(n−1) (−1)^k [−(1/(2(2x+1)))]_k ^(k+1) =−(1/2) Σ_(k=0) ^(n−1) (−1)^k { (1/(2k+3)) −(1/(2k+1))} =−(1/2)Σ_(k=0) ^(n−1) (((−1)^k )/(2k+3)) +(1/2)Σ_(k=0) ^(n−1) (((−1)^k )/(2k+1)) but Σ_(k=0) ^(n−1) (((−1)^k )/(2k+3)) = (1/3) +Σ_(k=1) ^(n−1) (((−1)^k )/(2k+3)) =_(k−1=p) (1/3) +Σ_(p=0) ^(n−2) (((−1)^(p+1) )/(2p+1)) =(1/3) − Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) ⇒ I_n = −(1/6) +(1/2) Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) +(1/2)Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) +(((−1)^(n−1) )/(2n−1)) I_n =−(1/6) +(((−1)^(n−1) )/(2n−1)) +Σ_(k=0) ^(n−2) (((−1)^k )/(2k+1)) 2)lim_(n→+∞) I_n =−(1/6) +Σ_(k=0) ^∞ (((−1)^k )/(2k+1)) =(π/4) −(1/6) .

$$\left.\mathrm{1}\right)\:{I}_{{n}} \:=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \left[−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \left\{\:\:\:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{3}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:+\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{3}} \\ $$$$=_{{k}−\mathrm{1}={p}} \:\:\frac{\mathrm{1}}{\mathrm{3}}\:+\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{2}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:−\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{2}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:\Rightarrow \\ $$$${I}_{{n}} \:=\:−\frac{\mathrm{1}}{\mathrm{6}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{2}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{2}} \frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}} \\ $$$${I}_{{n}} =−\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}{n}−\mathrm{1}}\:\:+\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{2}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){lim}_{{n}\rightarrow+\infty} \:\:{I}_{{n}} =−\frac{\mathrm{1}}{\mathrm{6}}\:+\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{6}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

[x]=0 1>x≥0 [x]=1 2>x≥1 [x]=2 3>x≥2 thus the value of (−1)^([x]) is either +1 or −1 ∫_0 ^1 (1/((2x+1)^2 ))dx+∫_1 ^2 ((−1)/((2x+1)^2 ))+∫_2 ^3 (1/((2x+1)^2 ))+... +∫_(n−1) ^n (((−1)^(n−1) )/((2x+1)^2 ))dx contd

$$\left[{x}\right]=\mathrm{0}\:\:\:\mathrm{1}>{x}\geqslant\mathrm{0} \\ $$$$\left[{x}\right]=\mathrm{1}\:\:\mathrm{2}>{x}\geqslant\mathrm{1} \\ $$$$\left[{x}\right]=\mathrm{2}\:\:\mathrm{3}>{x}\geqslant\mathrm{2} \\ $$$${thus}\:{the}\:{value}\:{of}\:\left(−\mathrm{1}\right)^{\left[{x}\right]} \:{is}\:{either}\:+\mathrm{1}\:{or}\:−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \frac{−\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }+\int_{\mathrm{2}} ^{\mathrm{3}} \frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }+… \\ $$$$+\int_{{n}−\mathrm{1}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${contd} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18