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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

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

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

$$ithink{it}^{\prime }s$$$$\underset{k=1}{\sum ^{2^n-1}}\frac{1}{k}$$

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

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{sir},\mathrm{I}\mathrm{understood}.$$

Commented by infinityaction last updated on 13/Oct/22

$$yessirpleasesolveit$$

Answered by mindispower last updated on 13/Oct/22

$$\frac{1}{2^n-1}⩽\frac{1}{2^{n-1}}\iff 2^n-1⩾2^{n-1}\iff 2^{n-1}⩾1true$$$$sincen⩾5$$$$S⩽\sum _{n⩾1}\frac{1}{2^{n-1}}=\underset{k=0}{\sum ^{n-1}}\frac{1}{2^k}=\frac{1-{\left(\frac{1}{2}\right)}^n}{\frac{1}{2}}=2.\frac{2^n-1}{2^n}⩽2$$$$\iff 1⩽s⩽2$$

Commented by mindispower last updated on 13/Oct/22

$$2⩽\frac{5}{2}⩽\frac{n}{2}$$

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