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Question Number 32098 by Tinkutara last updated on 19/Mar/18

Commented by ajfour last updated on 19/Mar/18

$$\frac{3L}{4}.$$

Commented by Tinkutara last updated on 19/Mar/18

Yes correct. Did you use torque balancing or COM concept?

Commented by ajfour last updated on 19/Mar/18

$$comconcept.$$

Commented by mrW2 last updated on 19/Mar/18

$$whatis{x}_{max}ifyoucantakeasmany$$$$blocksasyoulike?$$

Commented by mrW2 last updated on 20/Mar/18

Commented by mrW2 last updated on 20/Mar/18

$$x_n=\frac{L}{2}(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n})$$$$\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\mathrm{\infty }$$$$itmeanswithsufficientblocksone$$$$canreacheveryoverhang.$$

Commented by Tinkutara last updated on 20/Mar/18

@ajfour Sir please show COM concept. I can't solve by that method.

Commented by Tinkutara last updated on 20/Mar/18

$$@mrW2Sirhow$$$$\underset{n\to \mathrm{\infty }}{\lim }{x}_n=\mathrm{\infty }$$$$?$$

Commented by mrW2 last updated on 20/Mar/18

$$theharminicseriesisdivergent.$$$$x_{n+1}>x_n$$$$\Rightarrow \underset{n\to \mathrm{\infty }}{\lim }{x}_n=\mathrm{\infty }$$$$$$$$proof:$$$$assumeitconverges,i.e.$$$$S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}....$$$$S>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}....$$$$=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}....$$$$=\frac{1}{2}+(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....)$$$$\Rightarrow S>\frac{1}{2}+S$$$$thisiscontradiction.$$

Commented by Tinkutara last updated on 20/Mar/18

Thank you very much Sir! ��������

Commented by abdo imad last updated on 20/Mar/18

$$letput{H}_n=\sum _{k=1}^n\frac{1}{k}andprovethat{lim}_{n\to \mathrm{\infty }}{H}_n=+\mathrm{\infty }$$$$wehave{H}_n=1+\sum _{k=2}^n{\int }_{k-1}^k\frac{dt}{k}butk-1⩽t⩽k\Rightarrow$$$$\frac{1}{k}⩽\frac{1}{t}⩽\frac{1}{k-1}\Rightarrow \frac{1}{k}⩽{\int }_{k-1}^k\frac{dt}{t}⩽\frac{1}{k-1}\Rightarrow$$$$\sum _{k=2}^n\frac{1}{k}⩽\sum _{k=2}^n{\int }_{k-1}^k\frac{dt}{t}⩽\sum _{k=2}^n\frac{1}{k-1}\Rightarrow$$$$1+\sum _{k=2}^n\frac{1}{k}⩽1+\sum _{k=2}^n(ln\left(k\right)-ln(k-1))⩽1+\sum _{k=1}^{n-1}\frac{1}{k}\Rightarrow$$$$H_n⩽1+ln\left(n\right)⩽1+H_{n-1}\Rightarrow H_{n-1}⩾ln\left(n\right)\Rightarrow$$$$H_n⩾ln(n+1)\mathrm{\forall }n\in N^{★}but{lim}_{n\to \mathrm{\infty }}ln(n+1)=+\mathrm{\infty }\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}=+\mathrm{\infty }.alsowehavetherelation$$$$H_n-1⩽ln\left(n\right)⩽H_{n-1}forn⩾2.$$

Commented by Tinkutara last updated on 21/Mar/18

Thank you abdo imad Sir!

Commented by abdo imad last updated on 21/Mar/18

$$nevermindsir$$

Commented by mrW2 last updated on 22/Mar/18

$$thanks!$$$$nicetoknow:$$$$\ln (n+1)<H_n<\ln \left(n\right)+1$$

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