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Question Number 184062 by CrispyXYZ last updated on 02/Jan/23

$$\mathrm{Prove}\mathrm{that}$$ $$\underset{i=1}{\sum ^n}\frac{1}{\sqrt{i^2+i}}>\ln (n+1)$$

Answered by mr W last updated on 02/Jan/23

$$thingstoknow:$$ $$1)$$ $$\frac{a+b}{2}⩾\sqrt{ab}$$ $$\frac{1}{\sqrt{ab}}⩾\frac{2}{a+b}$$ $$2)$$ $$f\left(x\right)=2(\sqrt{x+1}-1)-\ln (x+1)$$ $$f\left(0\right)=0$$ $$f^{\prime }\left(x\right)=\frac{1}{\sqrt{x+1}}(1-\frac{1}{\sqrt{x+1}})>0$$ $$itmeansforx⩾0f\left(x\right)isstrictly$$ $$increasing,i.e.f\left(x\right)>0$$ $$\Rightarrow 2(\sqrt{x+1}-1)>\ln (x+1)$$ $$$$ $$\underset{i=1}{\sum ^n}\frac{1}{\sqrt{i^2+i}}$$ $$=\underset{i=1}{\sum ^n}\frac{1}{\sqrt{i(i+1)}}$$ $$>\underset{k=1}{\sum ^n}\left(\frac{2}{\sqrt{i}+\sqrt{i+1}}\right)see1)above$$ $$=2\underset{k=1}{\sum ^n}(\sqrt{i+1}-\sqrt{i})$$ $$=2(\sqrt{n+1}-1)$$ $$>\ln (n+1)✓see2)above$$

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