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Question Number 63101 by mathmax by abdo last updated on 29/Jun/19

$$calculateS=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+.....$$

Commented by mathmax by abdo last updated on 29/Jun/19

$$lettryanotherwaywehave\frac{d}{dx}ln(1+x)=\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{n}with\mid x\mid <1\Rightarrow$$$$ln(1+x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{n+1}}{n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}{x}^n}{n}\Rightarrow$$$${lim}_{x\to 1}ln(1+x)=ln\left(2\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...=$$$$=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+...=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+....=S\Rightarrow$$$$S=ln\left(2\right).$$

Answered by mr W last updated on 29/Jun/19

$$S=\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+.....>0$$$$S=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+$$$$S=(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...)-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)$$$$S=(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...)-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)$$$$S=(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...)-(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...)$$$$S=0?$$

Commented by MJS last updated on 29/Jun/19

$$\mathrm{I}\mathrm{seem}\mathrm{to}\mathrm{remember}\mathrm{that}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{allowed}$$$$\mathrm{to}\mathrm{rearrange}\mathrm{in}\mathrm{this}\mathrm{way}$$$$S=\frac{1}{2}+s\Rightarrow S>\frac{1}{2}$$$$S=\frac{7}{12}+s\Rightarrow S>\frac{7}{12}$$$$...$$

Commented by turbo msup by abdo last updated on 29/Jun/19

$$wehsveS=\mathrm{\Sigma }{a}_{i}and{a}_i>0how$$$$canSbe0....$$

Commented by turbo msup by abdo last updated on 29/Jun/19

$$wehsveS=\mathrm{\Sigma }{a}_{i}and{a}_i>0how$$$$canSbe0....$$

Commented by mathmax by abdo last updated on 29/Jun/19

$$sirmrwyourbeginingiscorrectbuttheanswerisnot0letcompletethe$$$$workwehaveS={lim}_{n\to +\mathrm{\infty }}(\sum _{k=0}^n\frac{1}{2k+1}-\frac{1}{2}\sum _{k=1}^n\frac{1}{k})$$$$wehave\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=0}^n\frac{1}{2k+1}=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{2n}+\frac{1}{2n+1}$$$$-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2n}=H_{2n+1}-\frac{1}{2}{H}_n\Rightarrow$$$$\sum _{k=0}^n\frac{1}{2k+1}-\frac{1}{2}\sum _{k=1}^n\frac{1}{k}=H_{2n+1}-H_n=ln(2n+1)+\gamma +o\left(\frac{1}{n}\right)-ln\left(n\right)-\gamma +o\left(\frac{1}{n}\right)$$$$=ln\left(\frac{2n+1}{n}\right)+o\left(\frac{1}{n}\right)\to ln\left(2\right)(n\to +\mathrm{\infty })so$$$$\frac{1}{1\times 2}+\frac{1}{3\times 4}+\frac{1}{5\times 6}+.....=ln\left(2\right).\left(\gamma istheconstantofeuler\right)$$

Commented by mr W last updated on 29/Jun/19

$$thankssir!ilearntsomethingnew.$$

Commented by mathmax by abdo last updated on 29/Jun/19

$$youarealwayswelcome.$$

Answered by MJS last updated on 29/Jun/19

$$\mathrm{it}\mathrm{seems}\mathrm{to}\mathrm{be}$$$$S=\ln 2$$$$\mathrm{but}\mathrm{I}\mathrm{cannot}\mathrm{properly}\mathrm{show}\mathrm{it}$$

Answered by naka3546 last updated on 29/Jun/19

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