Question Number 111622 by mathdave last updated on 04/Sep/20
![show that the close form of Σ_(k=0) ^∞ (Σ_(i=0) ^∞ [(((−1)^(k+i) )/((k+1)(k+2i+2)))])=(1/8)ln2](https://www.tinkutara.com/question/Q111622.png)
$${show}\:{that}\:{the}\:{close}\:{form}\:{of}\: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}+{i}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}{i}+\mathrm{2}\right)}\right]\right)=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln2} \\ $$
Answered by mathdave last updated on 04/Sep/20
![solution I=Σ_(k=0) ^∞ (Σ_(i=0) ^∞ [(((−1)^(i+k) )/(2i+1))((1/(k+1))−(1/(k+2i+2)))]) I=Σ_(k=0) ^∞ (Σ_(i=0) ^∞ [(((−1)^(i+k) )/((2i+1)(k+1)))−(((−1)^(i+k) )/((2i+1)(k+2i+2)))]) I=(Σ_(k=0) ^∞ (((−1)^k )/(k+1)))(Σ_(i=0) ^∞ (((−1)^i )/(2i+1)))−Σ_(k=0) ^∞ Σ_(i=0) ^∞ (((−1)^(i+k) )/((2i+1)(k+2i+2))) I=(Σ_(k=0) ^∞ (−1)^k ∫_0 ^1 y^k dy)(Σ_(i=0) ^∞ (−1)^i ∫_0 ^1 m^(2i) dm)−(Σ_(k=0) ^∞ (−1)^k ∫_0 ^1 u^((k+2i+1)) du)Σ_(i=0) ^∞ (((−1)^i )/(2i+1)) I=(∫_0 ^1 (dy/(1+y)))(∫_0 ^1 (dm/(1+m^2 )))−(Σ_(k=0) ^∞ (−1)^k ∫_0 ^1 u^k du)Σ_(i=0) ^∞ (((−1)^i u^(2i+1) )/(2i+1)) but tan^(−1) (u)=Σ_(i=0) ^∞ (((−1)^i u^(2i+1) )/(2i+1)) I=(ln2)(tan^(−1) (m))_0 ^1 −(∫_0 ^1 (du/(1+u)))tan^(−1) (u) I=(ln2)((π/4))−(∫_0 ^1 ((tan^(−1) u)/(1+u))du) using IBP I=(π/4)ln2−[(ln(1+u)tan^(−1) u]_0 ^1 +(∫_0 ^1 ((ln(1+u))/(1+u^2 ))du) I=(π/4)ln2−(π/4)ln2+∫_0 ^1 ((ln(1+u))/(1+u^2 ))du I=∫_0 ^1 ((ln(1+u))/(1+u^2 ))du (let u=tanx and du=(1+tan^2 x)dx I=∫_0 ^(π/4) ln(1+tanx)dx.........(1) using the property ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx I=∫_0 ^(π/4) ln[1+tan((π/4)−x)]dx= I=∫_0 ^(π/4) ln(1+((1−tanx)/(1+tanx)))dx I=∫_0 ^(π/4) (ln2−ln(1+tanx))dx........(2) adding (1) and (2) 2I=∫_0 ^(π/4) (ln(1+tanx)+ln2−ln(1+tanx))dx I=(1/2)∫_0 ^(π/4) ln2dx=(1/2)ln2∫_0 ^(π/4) dx=((ln2)/2)[x]_0 ^(π/4) I=(1/2)ln2×(π/4)=(1/8)ln2 ∵Σ_(k=0) ^∞ (Σ_(i=0) ^∞ [(((−1)^(i+k) )/((k+1)(k+2i+2)))])=(1/8)ln2 Q.E.D by mathdave(04/09/2020)](https://www.tinkutara.com/question/Q111776.png)
$${solution} \\ $$$${I}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{i}+{k}} }{\mathrm{2}{i}+\mathrm{1}}\left(\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{2}{i}+\mathrm{2}}\right)\right]\right) \\ $$$${I}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{i}+{k}} }{\left(\mathrm{2}{i}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)}−\frac{\left(−\mathrm{1}\right)^{{i}+{k}} }{\left(\mathrm{2}{i}+\mathrm{1}\right)\left({k}+\mathrm{2}{i}+\mathrm{2}\right)}\right]\right) \\ $$$${I}=\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\right)\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}} }{\mathrm{2}{i}+\mathrm{1}}\right)−\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}+{k}} }{\left(\mathrm{2}{i}+\mathrm{1}\right)\left({k}+\mathrm{2}{i}+\mathrm{2}\right)} \\ $$$${I}=\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{k}} {dy}\right)\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \int_{\mathrm{0}} ^{\mathrm{1}} {m}^{\mathrm{2}{i}} {dm}\right)−\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\left({k}+\mathrm{2}{i}+\mathrm{1}\right)} {du}\right)\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}} }{\mathrm{2}{i}+\mathrm{1}} \\ $$$${I}=\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dy}}{\mathrm{1}+{y}}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dm}}{\mathrm{1}+{m}^{\mathrm{2}} }\right)−\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{k}} {du}\right)\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}} {u}^{\mathrm{2}{i}+\mathrm{1}} }{\mathrm{2}{i}+\mathrm{1}} \\ $$$${but}\:\mathrm{tan}^{−\mathrm{1}} \left({u}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{i}} {u}^{\mathrm{2}{i}+\mathrm{1}} }{\mathrm{2}{i}+\mathrm{1}} \\ $$$${I}=\left(\mathrm{ln2}\right)\left(\mathrm{tan}^{−\mathrm{1}} \:\left({m}\right)\right)_{\mathrm{0}} ^{\mathrm{1}} −\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{du}}{\mathrm{1}+{u}}\right)\mathrm{tan}^{−\mathrm{1}} \left({u}\right) \\ $$$${I}=\left(\mathrm{ln2}\right)\left(\frac{\pi}{\mathrm{4}}\right)−\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} {u}}{\mathrm{1}+{u}}{du}\right) \\ $$$${using}\:{IBP} \\ $$$${I}=\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\left[\left(\mathrm{ln}\left(\mathrm{1}+{u}\right)\mathrm{tan}^{−\mathrm{1}} {u}\right]_{\mathrm{0}} ^{\mathrm{1}} +\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\right)\right. \\ $$$${I}=\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:\:\left({let}\:\:{u}=\mathrm{tan}{x}\:\:{and}\:{du}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right){dx}\right. \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tan}{x}\right){dx}………\left(\mathrm{1}\right)\:\:{using}\:{the}\:{property} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left[\mathrm{1}+\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]{dx}= \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}−\mathrm{tan}{x}}{\mathrm{1}+\mathrm{tan}{x}}\right){dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{ln2}−\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}{x}\right)\right){dx}……..\left(\mathrm{2}\right) \\ $$$${adding}\:\left(\mathrm{1}\right)\:\:{and}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}{x}\right)+\mathrm{ln2}−\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}{x}\right)\right){dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln2}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {dx}=\frac{\mathrm{ln2}}{\mathrm{2}}\left[{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln2}×\frac{\pi}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln2} \\ $$$$\because\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{i}+{k}} }{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}{i}+\mathrm{2}\right)}\right]\right)=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln2}\:\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave}\left(\mathrm{04}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$