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Question Number 147122 by mathdanisur last updated on 18/Jul/21

$$\underset{\mathit{n}\to \mathrm{\infty }}{lim}\underset{\mathit{k}=1}{\sum ^{\mathit{n}}}{2}^{\mathit{k}}\cdot {(\sqrt[2^{\mathit{k}}]{2}-1)}^2=?$$

Answered by Kamel last updated on 18/Jul/21

$$Sorryyouareright:\left(i^{\prime }mconsider{\left(\sqrt{2}\right)}^{\frac{1}{2^k}}\right)$$$$S_n=\underset{k=1}{\sum ^n}{2}^k{(\sqrt[2^k]{2}-1)}^2=\underset{k=1}{\sum ^n}{2}^k(2^{\frac{1}{2^{k-1}}}-2^{\frac{1}{2^k}+1}+1)$$$$=\underset{k=1}{\sum ^n}(2^{k+\frac{1}{2^{k-1}}}-2^{\frac{1}{2^k}+k+1})+2^{n+1}-1$$$$=2^2-2^{\frac{1}{2^1}+2}+2^{\frac{1}{2}+2}-2^{\frac{1}{2^2}+3}+2^{\frac{1}{2^2}+3}-2^{\frac{1}{2^3}+4}...-2^{\frac{1}{2^n}+n+1}+2^{n+1}-2$$$$=2-{22}^n(2^{\frac{1}{2^n}}-1)$$$$\text{Double subscripts: use braces to clarify}$$$$\therefore \underset{n\to +\mathrm{\infty }}{lim}\underset{k=1}{\sum ^n}{2}^k{(\sqrt[2^k]{2}-1)}^2=2-2Ln\left(2\right)$$$$\mathit{K}\mathit{A}\mathit{M}\mathit{E}\mathit{L}\mathit{B}\mathit{E}\mathit{N}\mathit{A}\mathit{I}\mathit{C}\mathit{H}\mathit{A}$$

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