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Question Number 184062 by CrispyXYZ last updated on 02/Jan/23
Prove that  Σ_(i=1) ^n  (1/( (√(i^2 +i)))) > ln(n+1)
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\:\sqrt{{i}^{\mathrm{2}} +{i}}}\:>\:\mathrm{ln}\left({n}+\mathrm{1}\right) \\ $$
Answered by mr W last updated on 02/Jan/23
things to know:  1)  ((a+b)/2)≥(√(ab))  (1/( (√(ab))))≥(2/(a+b))  2)  f(x)=2((√(x+1))−1)−ln (x+1)  f(0)=0  f′(x)=(1/( (√(x+1))))(1−(1/( (√(x+1)))))>0  it means for x≥0 f(x) is strictly  increasing,i.e. f(x)>0  ⇒2((√(x+1))−1)>ln (x+1)    Σ_(i=1) ^n (1/( (√(i^2 +i))))  =Σ_(i=1) ^n (1/( (√(i(i+1)))))  >Σ_(k=1) ^n ((2/( (√i)+(√(i+1)))))        see 1) above  =2Σ_(k=1) ^n ((√(i+1))−(√i))  =2((√(n+1))−1)  >ln (n+1) ✓              see 2) above
$${things}\:{to}\:{know}: \\ $$$$\left.\mathrm{1}\right) \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{ab}}}\geqslant\frac{\mathrm{2}}{{a}+{b}} \\ $$$$\left.\mathrm{2}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}\left(\sqrt{{x}+\mathrm{1}}−\mathrm{1}\right)−\mathrm{ln}\:\left({x}+\mathrm{1}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}}\right)>\mathrm{0} \\ $$$${it}\:{means}\:{for}\:{x}\geqslant\mathrm{0}\:{f}\left({x}\right)\:{is}\:{strictly} \\ $$$${increasing},{i}.{e}.\:{f}\left({x}\right)>\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left(\sqrt{{x}+\mathrm{1}}−\mathrm{1}\right)>\mathrm{ln}\:\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{i}^{\mathrm{2}} +{i}}} \\ $$$$=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{i}\left({i}+\mathrm{1}\right)}} \\ $$$$\left.>\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{\mathrm{2}}{\:\sqrt{{i}}+\sqrt{{i}+\mathrm{1}}}\right)\:\:\:\:\:\:\:\:{see}\:\mathrm{1}\right)\:{above} \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\sqrt{{i}+\mathrm{1}}−\sqrt{{i}}\right) \\ $$$$=\mathrm{2}\left(\sqrt{{n}+\mathrm{1}}−\mathrm{1}\right) \\ $$$$\left.>\mathrm{ln}\:\left({n}+\mathrm{1}\right)\:\checkmark\:\:\:\:\:\:\:\:\:\:\:\:\:\:{see}\:\mathrm{2}\right)\:{above} \\ $$

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