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Question Number 63574 by Tawa1 last updated on 05/Jul/19

$$\mathrm{prove}\mathrm{that}\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\mathrm{k}(2\mathrm{k}+1)}=2-2\ln \left(2\right)$$

Commented by mathmax by abdo last updated on 05/Jul/19

$$the{H}_{n}methodlet{S}_n=\sum _{k=1}^n\frac{1}{k(2k+1)}\Rightarrow \frac{1}{2}{S}_n=\sum _{k=1}^n\frac{1}{2k(2k+1)}$$$$=\sum _{k=1}^n\{\frac{1}{2k}-\frac{1}{2k+1}\}=\frac{1}{2}\sum _{k=1}^n\frac{1}{k}-\sum _{k=1}^n\frac{1}{2k+1}wehave$$$$\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=1}^n\frac{1}{2k+1}=\frac{1}{3}+\frac{1}{5}+......+\frac{1}{2n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2n}+\frac{1}{2n+1}$$$$-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2n}=H_{2n+1}-1-\frac{1}{2}{H}_n\Rightarrow$$$$\frac{1}{2}{S}_n=\frac{1}{2}{H}_n-H_{2n+1}+1+\frac{1}{2}{H}_n=H_n-H_{2n+1}+1$$$$=ln\left(n\right)+\gamma +o\left(\frac{1}{n}\right)-ln(2n+1)-\gamma +o\left(\frac{1}{n}\right)+1$$$$=ln\left(\frac{n}{2n+1}\right)+o\left(\frac{1}{n}\right)+1\Rightarrow S_n=2ln\left(\frac{n}{2n+1}\right)+2\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=2ln\left(\frac{1}{2}\right)+2=-2ln\left(2\right)+2.theresultisproved.$$

Commented by Prithwish sen last updated on 05/Jul/19

$$=\mathrm{\Sigma }[\frac{1}{\mathrm{k}}-\frac{2}{2\mathrm{k}+1}]$$$$=(1+\frac{1}{2}+\frac{1}{3}+..........\mathrm{\infty })-2(\frac{1}{3}+\frac{1}{5}+......\mathrm{\infty })$$$$=(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-.......\mathrm{\infty })$$$$=2-(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......\mathrm{\infty })$$$$=2-\ln \left(2\right)$$$$\mathrm{I}.{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{find}\mathrm{out}\mathrm{my}\mathrm{mistake}.\mathrm{Please}\mathrm{help}\mathrm{anyone}$$

Commented by Tawa1 last updated on 05/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sirs}$$

Commented by mathmax by abdo last updated on 05/Jul/19

$$methodofseriesletS=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(2n+1)}andS\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n+1}}{n(2n+1)}$$$$wehaveS=S\left(1\right)wehave\frac{dS}{dx}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n}}{n}=w\left(x^2\right)withw\left(t\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{t^n}{n}$$$$wehave{w}^{{}^{\prime }}\left(t\right)=\sum _{n=1}^{\mathrm{\infty }}{t}^{n-1}=\sum _{n=0}^{\mathrm{\infty }}{t}^n=\frac{1}{1-t}\Rightarrow w\left(t\right)=-ln\mid 1-t\mid +c$$$$c=w\left(0\right)=0\Rightarrow \frac{dS}{dx}\left(x\right)=-ln(1-x^2)(wetake-1<x<1)\Rightarrow$$$$S\left(x\right)=-{\int }_0^{x}ln(1-t^2)dt+\lambda (\lambda =S\left(0\right)=0)\Rightarrow S\left(x\right)=-{\int }_0^{x}ln(1-t^2)dt\Rightarrow$$$$S=-{\int }_0^{1}ln(1-t^2)dt=-{\int }_0^{1}ln(1-t)-{\int }_0^{1}ln(1+t)dt$$$${\int }_0^{1}ln(1-t)dt{=}_{1-t=u}{\int }_0^{1}ln\left(u\right)du={[uln\left(u\right)-u]}_0^1=-1$$$${\int }_0^{1}ln(1+t)dt{=}_{1+t=u}{\int }_1^{2}ln\left(u\right)du={[uln\left(u\right)-u]}_1^2=2ln\left(2\right)-2+1\Rightarrow$$$$S=1-2ln\left(2\right)+2-1=2-2ln\left(2\right).$$$$$$

Commented by Tawa1 last updated on 05/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 05/Jul/19

Commented by Tawa1 last updated on 05/Jul/19

$$\mathrm{Please}\mathrm{help}\mathrm{sir}$$

Commented by Tawa1 last updated on 05/Jul/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}$$

Commented by Prithwish sen last updated on 06/Jul/19

$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{answer}?$$

Commented by Tawa1 last updated on 06/Jul/19

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{answer}\mathrm{sir}.\mathrm{But}\mathrm{you}\mathrm{can}\mathrm{help}\mathrm{sir}.$$

Commented by Prithwish sen last updated on 06/Jul/19

$${\mathrm{a}}_{\mathrm{n}+1}=\sqrt{1+\mathrm{n}\sqrt{1+(\mathrm{n}-1)\sqrt{1+(\mathrm{n}-2)\sqrt{1+....}}}}$$$$\mathrm{Now},\mathrm{n}-1=\sqrt{{(\mathrm{n}-1)}^2}=\sqrt{1+\mathrm{n}(\mathrm{n}-2)}$$$$=\sqrt{1+\mathrm{n}\sqrt{{(\mathrm{n}-2)}^2}}=\sqrt{1+\mathrm{n}\sqrt{1+(\mathrm{n}-1)(\mathrm{n}-2)}}$$$$=\sqrt{1+\mathrm{n}\sqrt{1+(\mathrm{n}-1)\sqrt{1+(\mathrm{n}-2)\sqrt{1+..}}}}$$$$\mathrm{as}\mathrm{n}\to \mathrm{\infty }{\mathrm{a}}_{\mathrm{n}+1}\to (\mathrm{n}-1)\Rightarrow {\mathrm{a}}_{\mathrm{n}}\to (\mathrm{n}-2)\Rightarrow \frac{{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}}\to (1-\frac{2}{\mathrm{n}})\to 1$$$$\mathrm{\Omega }=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}{\left[\frac{1}{\mathrm{n}}\{\frac{1}{(\mathrm{n}+\frac{{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}})}+\frac{2}{(\mathrm{n}+2.\frac{{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}})}+.....+\frac{\mathrm{n}}{(\mathrm{n}+\mathrm{n}.\frac{{\mathrm{a}}_{\mathrm{n}}}{\mathrm{n}})}\}\right]}^{{\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1}$$$$=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}{\left[\frac{1}{\mathrm{n}}\{\frac{\mathrm{l}}{\mathrm{n}+1}+\frac{2}{\mathrm{n}+2}+..........+\frac{\mathrm{n}}{\mathrm{n}+\mathrm{n}}\}\right]}^{{\mathrm{n}}^{\frac{1}{\mathrm{n}}-1}}$$$$=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}{\left[\frac{1}{\mathrm{n}}\{\frac{\frac{1}{\mathrm{n}}}{1+\frac{1}{\mathrm{n}}}+\frac{\frac{2}{\mathrm{n}}}{1+\frac{2}{\mathrm{n}}}+.....+\frac{\frac{\mathrm{n}}{\mathrm{n}}}{1+\frac{\mathrm{n}}{\mathrm{n}}}\}\right]}^{{\mathrm{n}}^{\frac{1}{\mathrm{n}}-1}}$$$$\mathrm{taking}\log$$$$\ln \left(\mathrm{\Omega }\right)=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}({\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1)\ln \left[{\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{1}{\mathrm{n}}\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\frac{\frac{\mathrm{r}}{\mathrm{n}}}{1+\frac{\mathrm{r}}{\mathrm{n}}}\right]$$$$=\mathrm{li}\underset{\mathrm{n}\to \mathrm{\infty }}{\mathrm{m}}({\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1)\ln \left[{\int }_0^1\frac{\mathrm{xdx}}{1+\mathrm{x}}\right]$$$$={\lim }_{\mathrm{n}\to \mathrm{\infty }}({\mathrm{n}}^{\frac{1}{\mathrm{n}}}-1)\ln (1-\ln 2)$$$$\mathrm{Now}\mathrm{as}\mathrm{n}\to \mathrm{\infty }\ln \mathrm{\Omega }\to 0\Rightarrow \mathrm{\Omega }\to 1$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{whether}\mathrm{it}\mathrm{is}\mathrm{corrector}\mathrm{not}.\mathrm{Please}$$$$\mathrm{give}\mathrm{your}\mathrm{valuable}\mathrm{suggestion}.$$

Commented by Prithwish sen last updated on 06/Jul/19

$$\mathrm{thanks}\mathrm{abdo}\mathrm{sir}$$

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