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Question Number 227853 by MrAjder last updated on 22/Feb/26
prove:Σ_(i=1) ^n (ln((i+1)/i))^2 <(n/(n+1))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{ln}\frac{{i}+\mathrm{1}}{{i}}\right)^{\mathrm{2}} <\frac{{n}}{{n}+\mathrm{1}} \\ $$
Answered by MrAjder last updated on 22/Feb/26
∀n∈N^+ ,S_n =Σ_(i=1) ^n (ln((i+1)/i))^2 <(n/(n+1)) Lemma:∀t>0,ln(1+t)<(t/( (√(1+t)))) Prove:let g(t)=(t/( (√(1+t))))−ln(1+t),g(0)=0 g′(t)=((1+(t/2))/((1−t)^(3/2) ))−(1/(1−t))=((1+(t/2)−(√(1−t)))/((1+t)^(3/2) )) let u=(√(1+t))>0,1+(t/2)=((u^2 +1)/2),g′(t)=((((u^2 +1)/2)−u)/u^3 )=(((u−1)^2 )/(2u^3 ))≥0 ∴g(t)≥g(0)=0,∀t>0,ln(1+t)<(t/( (√(1+t)))) ∀m∈N^+ ,t=(1/m),ln(1+(1/m))<((1/m)/( (√(1+1/m))))=(1/( (√(m(m+1))))) ⇒(ln(1+(1/m)))^2 <(1/(m(m+1))) (∗) P(1):S_1 =(ln 2)^2 →^((∗)m=1) <(1/(1∙2))=(1/2) let P(n):S_n <(n/(n+1))If it gets established, then S_(n+1) =S_n +(ln(1+(1/(n+1))))^2 →^((∗)m=n+1) <(n/(n+1))+(1/((n+1)(n+2))) (n/(n+1))+(1/((n+1)(n+2)))=((n(n+2)+1)/((n+1)(n+2)))=(((n+1)^2 )/((n+1)(n+2)))=((n+1)/(n+2)) ∴S_(n+1) <((n+1)/(n+2)),That is P(n+1) is true By induction ∀n∈N^+ ,S_n <(n/(n+1))
$$\forall{n}\in\mathbb{N}^{+} ,{S}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{ln}\frac{{i}+\mathrm{1}}{{i}}\right)^{\mathrm{2}} <\frac{{n}}{{n}+\mathrm{1}} \\ $$$$\mathrm{Lemma}:\forall{t}>\mathrm{0},\mathrm{ln}\left(\mathrm{1}+{t}\right)<\frac{{t}}{\:\sqrt{\mathrm{1}+{t}}} \\ $$$$\mathrm{Prove}:\mathrm{let}\:{g}\left({t}\right)=\frac{{t}}{\:\sqrt{\mathrm{1}+{t}}}−\mathrm{ln}\left(\mathrm{1}+{t}\right),{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${g}'\left({t}\right)=\frac{\mathrm{1}+\frac{{t}}{\mathrm{2}}}{\left(\mathrm{1}−{t}\right)^{\mathrm{3}/\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}−{t}}=\frac{\mathrm{1}+\frac{{t}}{\mathrm{2}}−\sqrt{\mathrm{1}−{t}}}{\left(\mathrm{1}+{t}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$\mathrm{let}\:{u}=\sqrt{\mathrm{1}+{t}}>\mathrm{0},\mathrm{1}+\frac{{t}}{\mathrm{2}}=\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}},{g}'\left({t}\right)=\frac{\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}−{u}}{{u}^{\mathrm{3}} }=\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{u}^{\mathrm{3}} }\geq\mathrm{0} \\ $$$$\therefore{g}\left({t}\right)\geq{g}\left(\mathrm{0}\right)=\mathrm{0},\forall{t}>\mathrm{0},\mathrm{ln}\left(\mathrm{1}+{t}\right)<\frac{{t}}{\:\sqrt{\mathrm{1}+{t}}} \\ $$$$\forall{m}\in\mathbb{N}^{+} ,{t}=\frac{\mathrm{1}}{{m}},\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)<\frac{\mathrm{1}/{m}}{\:\sqrt{\mathrm{1}+\mathrm{1}/{m}}}=\frac{\mathrm{1}}{\:\sqrt{{m}\left({m}+\mathrm{1}\right)}} \\ $$$$\Rightarrow\left(\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)\right)^{\mathrm{2}} <\frac{\mathrm{1}}{{m}\left({m}+\mathrm{1}\right)}\:\left(\ast\right) \\ $$$${P}\left(\mathrm{1}\right):{S}_{\mathrm{1}} =\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} \overset{\left(\ast\right){m}=\mathrm{1}} {\rightarrow}<\frac{\mathrm{1}}{\mathrm{1}\centerdot\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{let}\:{P}\left({n}\right):{S}_{{n}} <\frac{{n}}{{n}+\mathrm{1}}\mathrm{If}\:\mathrm{it}\:\mathrm{gets}\:\mathrm{established},\:\mathrm{then} \\ $$$${S}_{{n}+\mathrm{1}} ={S}_{{n}} +\left(\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\right)^{\mathrm{2}} \overset{\left(\ast\right){m}={n}+\mathrm{1}} {\rightarrow}<\frac{{n}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$$$\frac{{n}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{{n}\left({n}+\mathrm{2}\right)+\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}=\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}} \\ $$$$\therefore{S}_{{n}+\mathrm{1}} <\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}},\mathrm{That}\:\mathrm{is}\:{P}\left({n}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{By}\:\mathrm{induction}\:\forall{n}\in\mathbb{N}^{+} ,{S}_{{n}} <\frac{{n}}{{n}+\mathrm{1}} \\ $$

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