Prove by the PMI, that

1/(n+1) + 1/(n+2) + .........+ 1/2n > 13/24 for all n > 1.

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{P}\left(\mathrm{n}\right): \frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}+......+\frac{1}{2\mathrm{n}}>\frac{13}{24} \mathrm{for} \mathrm{all} \mathrm{n}>1 \\ \mathbf{Step}\mathbf{ }\mathbf{1}\mathbf{:}\mathbf{ }\mathbf{Base}\mathbf{ }\mathbf{Case} \\ \mathrm{For} \mathrm{n}=2 \\ \frac{1}{2+1}+\frac{1}{2+2}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}>\frac{13}{24} \\ \mathrm{So} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=2. \\ \mathbf{Step}\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{ }\mathbf{Inductive}\mathbf{ }\mathbf{Hypothesis} \\ \mathrm{Let} \mathrm{P}\left(\mathrm{n}\right) \mathrm{be} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k} \mathrm{i}.\mathrm{e}. \\ \frac{1}{\mathrm{k}+1}+\frac{1}{\mathrm{k}+2}+.....+\frac{1}{2\mathrm{k}}>\frac{13}{24} \\ \mathbf{Step}\mathbf{ }\mathbf{3}\mathbf{:}\mathbf{ }\mathbf{Inductive}\mathbf{ }\mathbf{Case} \\ \mathrm{Consider} \\ {\mathrm{P}}_{\mathrm{k}+1}=\frac{1}{\left(\mathrm{k}+1\right)+1}+\frac{1}{\left(\mathrm{k}+1\right)+2}+......+\frac{1}{2\left(\mathrm{k}+1\right)-2}\frac{1}{2\left(\mathrm{k}+1\right)-1}+\frac{1}{2\left(\mathrm{k}+1\right)} \\ =\frac{1}{\mathrm{k}+2}+\frac{1}{\mathrm{k}+3}+.....+\frac{1}{2\mathrm{k}}+\frac{1}{2\mathrm{k}+1}+\frac{1}{2\left(\mathrm{k}+1\right)} \\ =-\frac{1}{\mathrm{k}+1}+\left(\frac{1}{\mathrm{k}+1}+\frac{1}{\mathrm{k}+2}+\frac{1}{\mathrm{k}+3}+.....+\frac{1}{2\mathrm{k}}\right)+\frac{1}{2\mathrm{k}+1}+\frac{1}{2\left(\mathrm{k}+1\right)} \\ \mathrm{Using} \mathrm{result} \mathrm{of} \mathrm{inductive} \mathrm{hypothesis} \mathrm{we} \mathrm{get}, \\ {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\frac{1}{2\mathrm{k}+1}+\frac{1}{2\left(\mathrm{k}+1\right)}-\frac{1}{\mathrm{k}+1} \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\frac{1}{2\mathrm{k}+1}+\frac{1}{\left(\mathrm{k}+1\right)}\left(\frac{1}{2}-1\right) \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\frac{1}{2\mathrm{k}+1}-\frac{1}{2\left(\mathrm{k}+1\right)} \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\frac{2\left(\mathrm{k}+1\right)-\left(2\mathrm{k}+1\right)}{\left(2\mathrm{k}+1\right)2\left(\mathrm{k}+1\right)} \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\frac{2\mathrm{k}+2-2\mathrm{k}-1}{2\left(2\mathrm{k}+1\right)\left(\mathrm{k}+1\right)} \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\frac{1}{2\left(2\mathrm{k}+1\right)\left(\mathrm{k}+1\right)} \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24}+\mathrm{Positive} \mathrm{Number} \\ \Rightarrow {\mathrm{P}}_{\mathrm{k}+1}>\frac{13}{24} \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k}+1. \\ \mathbf{So}\mathbf{ }\mathbf{P}\left(\mathbf{n}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{true}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{n}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{assumption}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{truth}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{n}\mathbf{=}\mathbf{k}\mathbf{ }\mathbf{follows}\mathbf{ }\mathbf{that}\mathbf{ }\mathbf{P}\left(\mathbf{n}\right)\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{n}\mathbf{=}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{also}\mathbf{ }\mathbf{true}\mathbf{.} \\ \mathbf{Hence}\mathbf{ }\mathbf{by}\mathbf{ }\mathbf{Principle}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Mathematical}\mathbf{ }\mathbf{Induction}\mathbf{ }\mathbf{P}\left(\mathbf{n}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{true}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{all}\mathbf{ }\mathbf{n}\mathbf{>}\mathbf{1} \end{gathered}$$

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