Question-8252

Question Number 8252 by 8168 last updated on 04/Oct/16

Answered by Yozzias last updated on 04/Oct/16

6+4+(8/3)+((16)/9)+... =6+((2^2 /3^0 )+(2^3 /3^1 )+(2^4 /3^2 )+...+(2^(n+1) /3^(n−1) )+...) =6+Σ_(n=1) ^∞ (2^(n+1) /3^(n−1) ) =6+(2/3^(−1) )Σ_(n=1) ^∞ ((2/3))^n =6+6×((2/3)/(1−(2/3))) =6+6×((2/3)/(1/3)) 6+4+(8/3)+((16)/9)+...=18 The sum is convergent and hence bounded and monotone. For s(n)=6+6Σ_(r=1) ^n ((2/3))^r ⇒s(n+1)−s(n)=6((2/3))^(n+1) >0 ⇒s(n+1)>s(n)⇒s(n) is an increasing sequence of partial sums. Since s(n) is increasing and lim_(n→∞) s(n)=18 then s(n) is bounded above and s(n)≤18 ∀n∈N.

$$\mathrm{6}+\mathrm{4}+\frac{\mathrm{8}}{\mathrm{3}}+\frac{\mathrm{16}}{\mathrm{9}}+… \\ $$$$=\mathrm{6}+\left(\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{0}} }+\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{1}} }+\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }{\mathrm{3}^{\mathrm{n}−\mathrm{1}} }+…\right) \\ $$$$=\mathrm{6}+\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{n}+\mathrm{1}} }{\mathrm{3}^{\mathrm{n}−\mathrm{1}} } \\ $$$$=\mathrm{6}+\frac{\mathrm{2}}{\mathrm{3}^{−\mathrm{1}} }\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}} \\ $$$$=\mathrm{6}+\mathrm{6}×\frac{\mathrm{2}/\mathrm{3}}{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$=\mathrm{6}+\mathrm{6}×\frac{\mathrm{2}/\mathrm{3}}{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{6}+\mathrm{4}+\frac{\mathrm{8}}{\mathrm{3}}+\frac{\mathrm{16}}{\mathrm{9}}+…=\mathrm{18} \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\mathrm{hence} \\ $$$$\mathrm{bounded}\:\mathrm{and}\:\mathrm{monotone}. \\ $$$$\mathrm{For}\:\mathrm{s}\left(\mathrm{n}\right)=\mathrm{6}+\mathrm{6}\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{r}} \\ $$$$\Rightarrow\mathrm{s}\left(\mathrm{n}+\mathrm{1}\right)−\mathrm{s}\left(\mathrm{n}\right)=\mathrm{6}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{n}+\mathrm{1}} >\mathrm{0} \\ $$$$\Rightarrow\mathrm{s}\left(\mathrm{n}+\mathrm{1}\right)>\mathrm{s}\left(\mathrm{n}\right)\Rightarrow\mathrm{s}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{increasing} \\ $$$$\mathrm{sequence}\:\mathrm{of}\:\mathrm{partial}\:\mathrm{sums}. \\ $$$$\mathrm{Since}\:\mathrm{s}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{and}\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}s}\left(\mathrm{n}\right)=\mathrm{18} \\ $$$$\mathrm{then}\:\mathrm{s}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{bounded}\:\mathrm{above}\:\mathrm{and} \\ $$$$\mathrm{s}\left(\mathrm{n}\right)\leqslant\mathrm{18}\:\forall\mathrm{n}\in\mathbb{N}. \\ $$$$ \\ $$

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