Menu Close

prove-that-k-1-n-1-1-k-gt-1-k-1-n-1-k-

Tinku Tara 0 Comments
Pin




Question Number 43001 by abdo.msup.com last updated on 06/Sep/18
prove that Π_(k=1) ^n (1+(1/k))>1+Σ_(k=1) ^n (1/k)
$${prove}\:{that}\:\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)>\mathrm{1}+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18
LHS (1+(1/1))(1+(1/2))(1+(1/3))..(1+(1/n)) let analysis (1+(1/1))(1+(1/2))=1+1((1/1)+(1/2))+((1/1)×(1/2)) (1+(1/1))(1+(1/2))(1+(1/3))= =(1+(1/3))(1+(1/1)+(1/2)+(1/(1×2))) =1+(1/1)+(1/2)+(1/(1×2))+(1/3)+(1/(1×3))+(1/(2×3))+(1/(1×2×3)) =1+(1/1)+(1/2)+(1/3)+(1/(1×2))+(1/(1×3))+(1/(2×3))+(1/(1×2×3)) thus Π_(k=1) ^3 (1+(1/k))>1+(1/1)+(1/2)+(1/3) Π_(k=1) ^3 (1+(1/k))>1+Σ_(k=1) ^3 (1/k) soΠ_(k=1) ^n (1+(1/k))>1+Σ_(k=1) ^n (1/k) proved
$${LHS} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)..\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right) \\ $$$${let}\:{analysis} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1}+\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)= \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}} \\ $$$${thus}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)>\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)>\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\frac{\mathrm{1}}{{k}} \\ $$$${so}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)>\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:{proved} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *