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Question Number 187085 by Rupesh123 last updated on 13/Feb/23

Answered by witcher3 last updated on 14/Feb/23

$${\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}$$$${\mathrm{S}}_{\mathrm{n}+1}-{\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+1}}\frac{1}{\mathrm{n}+1+\mathrm{k}}-\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}+\mathrm{k}}$$$$=\frac{1}{2\mathrm{n}+1}+\frac{1}{2\mathrm{n}+2}-\frac{1}{\mathrm{n}}$$$$=\frac{(2\mathrm{n}+2)\mathrm{n}+\mathrm{n}(2\mathrm{n}+1)-(2\mathrm{n}+2)(2\mathrm{n}+1)}{2(\mathrm{n}+1)\mathrm{n}(2\mathrm{n}+1)}$$$$=\frac{-3\mathrm{n}-2}{2\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}<0$$$${\mathrm{s}}_{\mathrm{n}}⩽\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{n}}=1$$$${\mathrm{s}}_{\mathrm{n}}\mathrm{cv}$$$$\left(2\right){\mathrm{T}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\sqrt{{2\mathrm{n}}^2+\mathrm{k}}}$$$$\mathrm{\forall }\mathrm{k}\in [1,\mathrm{n}]\frac{1}{\sqrt{{2\mathrm{n}}^2+\mathrm{n}}}⩽\frac{1}{\sqrt{{2\mathrm{n}}^2+\mathrm{k}}}⩽\frac{1}{\sqrt{{2\mathrm{n}}^2+1}}$$$$\Rightarrow \frac{\mathrm{n}}{\sqrt{{2\mathrm{n}}^2+\mathrm{n}}}\to \frac{1}{\sqrt{\mathrm{n}}}⩽{\mathrm{T}}_{\mathrm{n}}⩽\frac{\mathrm{n}}{\sqrt{{2\mathrm{n}}^2+1}}\to \frac{1}{\sqrt{2}}$$$$\text{Double subscripts: use braces to clarify}$$$$$$

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