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Question Number 178073 by infinityaction last updated on 12/Oct/22

Commented by Rasheed.Sindhi last updated on 12/Oct/22

(1/(2^1 −1))+(1/(2^2 −1))+(1/(2^3 −1))+...+(1/(2^n −1)) (1/(2−1))+(1/(4−1))+(1/(8−1))+...+(1/(2^n −1)) 1+(1/3)+(1/7)+...+(1/(2^n −1)) ?

$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} −\mathrm{1}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}−\mathrm{1}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{7}}+...+\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}}\:\:\:\:? \\ $$

Commented by mr W last updated on 12/Oct/22

i think it′s Σ_(k=1) ^(2^n −1) (1/k)

$${i}\:{think}\:{it}'{s} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}^{{n}} −\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$

Commented by Rasheed.Sindhi last updated on 12/Oct/22

Thanks sir, I understood.

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}},\:\mathrm{I}\:\mathrm{understood}. \\ $$

Commented by infinityaction last updated on 13/Oct/22

yes sir please solve it

$${yes}\:{sir}\:{please}\:{solve}\:{it}\: \\ $$

Answered by mindispower last updated on 13/Oct/22

(1/(2^n −1))≤(1/2^(n−1) )⇔2^n −1≥2^(n−1) ⇔2^(n−1) ≥1 true since n≥5 S≤Σ_(n≥1) (1/2^(n−1) )=Σ_(k=0) ^(n−1) (1/2^k )=((1−((1/2))^n )/(1/2))=2.((2^n −1)/2^n )≤2 ⇔ 1≤s≤2

$$\frac{\mathrm{1}}{\mathrm{2}^{{n}} −\mathrm{1}}\leqslant\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\Leftrightarrow\mathrm{2}^{{n}} −\mathrm{1}\geqslant\mathrm{2}^{{n}−\mathrm{1}} \Leftrightarrow\mathrm{2}^{{n}−\mathrm{1}} \geqslant\mathrm{1}\:{true} \\ $$$${since}\:{n}\geqslant\mathrm{5} \\ $$$${S}\leqslant\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{k}} }=\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} }{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2}.\frac{\mathrm{2}^{{n}} −\mathrm{1}}{\mathrm{2}^{{n}} }\leqslant\mathrm{2} \\ $$$$\Leftrightarrow\:\mathrm{1}\leqslant{s}\leqslant\mathrm{2} \\ $$

Commented by mindispower last updated on 13/Oct/22

2≤(5/2)≤(n/2)

$$\mathrm{2}\leqslant\frac{\mathrm{5}}{\mathrm{2}}\leqslant\frac{{n}}{\mathrm{2}} \\ $$

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