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Question Number 13295 by Tinkutara last updated on 17/May/17

For n ∈ N, n > 1, show that  (1/n) + (1/(n + 1)) + (1/(n + 2)) + ... + (1/n^2 ) > 1

$$\mathrm{For}\:{n}\:\in\:\mathbb{N},\:{n}\:>\:\mathrm{1},\:\mathrm{show}\:\mathrm{that} \\ $$ $$\frac{\mathrm{1}}{{n}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{2}}\:+\:...\:+\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:>\:\mathrm{1} \\ $$

Commented byprakash jain last updated on 17/May/17

n=2 and 3 can be checked  with hand calculation  n≥4  n^2 ≥4n  (1/n)+(1/(n+1))+(1/(n+2))+....+(1/n^2 )  >(1/n)+(1/(n+1))+.....+(1/(4n−1))  =(1/n)+(1/(n+1))+..+(1/(2n−1))          + (1/(2n))+(1/(2n+1))+...+(1/(3n−1))           +(1/(3n))+(1/(3n−1))+....+(1/(4n−1))  >(1/(2n−1))+...+(1/(2n−1))         +(1/(3n−1))+....+(1/(3n−1))          +(1/(4n−1))+...+(1/(4n−1))  =(n/(2n−1))+(n/(3n−1))+(n/(4n−1))  >(1/2)+(1/3)+(1/4)>1

$${n}=\mathrm{2}\:{and}\:\mathrm{3}\:{can}\:{be}\:{checked} \\ $$ $${with}\:{hand}\:{calculation} \\ $$ $${n}\geqslant\mathrm{4} \\ $$ $${n}^{\mathrm{2}} \geqslant\mathrm{4}{n} \\ $$ $$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{2}}+....+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$ $$>\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+.....+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}} \\ $$ $$=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+..+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$ $$\:\:\:\:\:\:\:\:+\:\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}+...+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}} \\ $$ $$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{3}{n}}+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}+....+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}} \\ $$ $$>\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}+...+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$ $$\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}+....+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}} \\ $$ $$\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}}+...+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}} \\ $$ $$=\frac{{n}}{\mathrm{2}{n}−\mathrm{1}}+\frac{{n}}{\mathrm{3}{n}−\mathrm{1}}+\frac{{n}}{\mathrm{4}{n}−\mathrm{1}} \\ $$ $$>\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}>\mathrm{1} \\ $$

Commented byTinkutara last updated on 18/May/17

Prove in general for any n ∈ N.

$$\mathrm{Prove}\:\mathrm{in}\:\mathrm{general}\:\mathrm{for}\:\mathrm{any}\:{n}\:\in\:\mathbb{N}. \\ $$

Commented byprakash jain last updated on 18/May/17

The above proves for all n∈N

$$\mathrm{The}\:\mathrm{above}\:\mathrm{proves}\:\mathrm{for}\:\mathrm{all}\:{n}\in\mathbb{N} \\ $$

Answered by mrW1 last updated on 18/May/17

Just to play......    There are n^2 −n+1 terms in total.  n^2 −n+1=(n−1)n+1  ⇒we can sort the terms in n groups:  group 1 with n terms:  S_1 =(1/n) + (1/(n + 1)) + (1/(n + 2)) + ... +(1/(2n−1))  group 2 with n terms:  S_2 =(1/(2n)) + (1/(2n + 1)) + (1/(2n + 2)) + ... +(1/(3n−1))  ......  group k with n terms:  S_k =(1/(kn)) + (1/(kn + 1)) + (1/(kn + 2)) + ... +(1/((k+1)n−1))  ......  group n−1 with n terms:  S_(n−1) =(1/((n−1)n)) + (1/((n−1)n + 1)) + (1/((n−1)n + 2)) + ... +(1/(n×n−1))  group n with 1 term:  S_n =(1/n^2 )    sum of terms in group k:  S_k =(1/(kn)) + (1/(kn + 1)) + (1/(kn + 2)) + ... +(1/((k+1)n−1))  >n×(1/((k+1)n−1))>(n/((k+1)n))=(1/(k+1))    sum of all terms:  S=Σ_(k=1) ^n S_k =Σ_(k=1) ^(n−1) S_k +S_n   >Σ_(k=1) ^(n−1) (1/(k+1))+(1/n^2 )  =(1/2)+(1/3)+(1/4)+∙∙∙+(1/n)+(1/n^2 )  =1+(1/2)+(1/3)+(1/4)+∙∙∙+(1/n)+(1/n^2 )−1  =Σ_(k=1) ^n (1/k)+(1/n^2 )−1  for big number n:  S>ln n+γ+(1/(2n))−(1/(12n^2 ))+(1/n^2 )−1  with γ=0.57721...  S>ln n+γ+(1/(2n))+((11)/(12n^2 ))−1

$${Just}\:{to}\:{play}...... \\ $$ $$ \\ $$ $${There}\:{are}\:{n}^{\mathrm{2}} −{n}+\mathrm{1}\:{terms}\:{in}\:{total}. \\ $$ $${n}^{\mathrm{2}} −{n}+\mathrm{1}=\left({n}−\mathrm{1}\right){n}+\mathrm{1} \\ $$ $$\Rightarrow{we}\:{can}\:{sort}\:{the}\:{terms}\:{in}\:{n}\:{groups}: \\ $$ $${group}\:\mathrm{1}\:{with}\:{n}\:{terms}: \\ $$ $${S}_{\mathrm{1}} =\frac{\mathrm{1}}{{n}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{n}\:+\:\mathrm{2}}\:+\:...\:+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$ $${group}\:\mathrm{2}\:{with}\:{n}\:{terms}: \\ $$ $${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}{n}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{n}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{n}\:+\:\mathrm{2}}\:+\:...\:+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}} \\ $$ $$...... \\ $$ $${group}\:{k}\:{with}\:{n}\:{terms}: \\ $$ $${S}_{{k}} =\frac{\mathrm{1}}{{kn}}\:+\:\frac{\mathrm{1}}{{kn}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{kn}\:+\:\mathrm{2}}\:+\:...\:+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){n}−\mathrm{1}} \\ $$ $$...... \\ $$ $${group}\:{n}−\mathrm{1}\:{with}\:{n}\:{terms}: \\ $$ $${S}_{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}}\:+\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right){n}\:+\:\mathrm{2}}\:+\:...\:+\frac{\mathrm{1}}{{n}×{n}−\mathrm{1}} \\ $$ $${group}\:{n}\:{with}\:\mathrm{1}\:{term}: \\ $$ $${S}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$ $$ \\ $$ $${sum}\:{of}\:{terms}\:{in}\:{group}\:{k}: \\ $$ $${S}_{{k}} =\frac{\mathrm{1}}{{kn}}\:+\:\frac{\mathrm{1}}{{kn}\:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{{kn}\:+\:\mathrm{2}}\:+\:...\:+\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){n}−\mathrm{1}} \\ $$ $$>{n}×\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right){n}−\mathrm{1}}>\frac{{n}}{\left({k}+\mathrm{1}\right){n}}=\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$ $$ \\ $$ $${sum}\:{of}\:{all}\:{terms}: \\ $$ $${S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{S}_{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{S}_{{k}} +{S}_{{n}} \\ $$ $$>\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}+\mathrm{1}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$ $$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1} \\ $$ $$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1} \\ $$ $${for}\:{big}\:{number}\:{n}: \\ $$ $${S}>\mathrm{ln}\:{n}+\gamma+\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{12}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}\:\:{with}\:\gamma=\mathrm{0}.\mathrm{57721}... \\ $$ $${S}>\mathrm{ln}\:{n}+\gamma+\frac{\mathrm{1}}{\mathrm{2}{n}}+\frac{\mathrm{11}}{\mathrm{12}{n}^{\mathrm{2}} }−\mathrm{1} \\ $$

Commented byprakash jain last updated on 18/May/17

γ=lim_(n→∞) [Σ_(i=1) ^n (1/i)−ln n]

$$\gamma=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{i}}−\mathrm{ln}\:{n}\right] \\ $$

Commented bymrW1 last updated on 18/May/17

S>Σ_(k=1) ^n (1/k)+(1/n^2 )−1>n((1/(n!)))^(1/n) +(1/n^2 )−1=(n/(^n (√(n!))))+(1/n^2 )−1

$${S}>\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}>{n}\left(\frac{\mathrm{1}}{{n}!}\right)^{\frac{\mathrm{1}}{{n}}} +\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}=\frac{{n}}{\:^{{n}} \sqrt{{n}!}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1} \\ $$

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