now i understand why Chinese always top one in IMO

Henry_2001 wrote: A sequence $\{a_n\}$ is defined recursively by $a_1=\frac{1}{2}, $ and for $n\ge 2,$ $0<a_n\leq a_{n-1}$ and \[a_n^2(a_{n-1}+1)+a_{n-1}^2(a_n+1)-2a_na_{n-1}(a_na_{n-1}+a_n+1)=0.\]$(1)$ Determine the general formula of the sequence $\{a_n\};$ We get $a_2=\frac 13$ and, from there, we have a quadratic whose two roots are $a_n=a_{n-1}\frac{1+a_{n-1}}{1-a_{n-1}-2a_{n-1}^2}$ or $a_n=\frac{a_{n-1}}{a_{n-1}+1}$ And so (since $a_n\le a_{n-1}$ : $a_n=\frac{a_{n-1}}{a_{n-1}+1}$, which is $\frac 1{a_n}=1+\frac 1{a_{n-1}}$ And so $\boxed{a_n=\frac 1{n+1}}$

Henry_2001 wrote: $(2)$ Let $S_n=a_1+\cdots+a_n.$ Prove that for $n\ge 1,$ $\ln\left(\frac{n}{2}+1\right)<S_n<\ln(n+1).$ Upper bound : We have $\frac 1{k+1}<\ln(k+1)-\ln k$ and so $S_n=\sum_{k=1}^n\frac 1{k+1}<\sum_{k=1}^n(\ln(k+1)-\ln k$ $=\ln (n+1)$ Lower bound : We have $\frac 1{k+1}>\ln(k+2)-\ln(k+1)$ and so $S_n=\sum_{k=1}^n\frac 1{k+1}<\sum_{k=1}^n(\ln(k+2)-\ln (k+1)$ $=\ln (n+2)-\ln 2=\ln(\frac n2+1)$

laikhanhhoang_3011 wrote: now i understand why Chinese always top one in IMO I still don't but thanks anyways.

starchan wrote: laikhanhhoang_3011 wrote: now i understand why Chinese always top one in IMO I still don't but thanks anyways. it's a popular joke in my country

pco wrote: Henry_2001 wrote: A sequence $\{a_n\}$ is defined recursively by $a_1=\frac{1}{2}, $ and for $n\ge 2,$ $0<a_n\leq a_{n-1}$ and \[a_n^2(a_{n-1}+1)+a_{n-1}^2(a_n+1)-2a_na_{n-1}(a_na_{n-1}+a_n+1)=0.\]$(1)$ Determine the general formula of the sequence $\{a_n\};$ We get $a_2=\frac 13$ and, from there, we have a quadratic whose two roots are $a_n=a_{n-1}\frac{1+a_{n-1}}{1-a_{n-1}-2a_{n-1}^2}$ or $a_n=\frac{a_{n-1}}{a_{n-1}+1}$ And so (since $a_n\le a_{n-1}$ : $a_n=\frac{a_{n-1}}{a_{n-1}+1}$, which is $\frac 1{a_n}=1+\frac 1{a_{n-1}}$ And so $\boxed{a_n=\frac 1{n+1}}$ How does you find $a_n$ : $a_n=a_{n-1}\frac{1+a_{n-1}}{1-a_{n-1}-2a_{n-1}^2}$ or $a_n=\frac{a_{n-1}}{a_{n-1}+1}$ , can you explain me ?

@above $a_n=\frac{a_{n-1}}{a_{n-1}+1}$ is a solution to the quadratic equation with $a_n<a_{n-1}$

bebeevan wrote: How does you find $a_n$ : $a_n=a_{n-1}\frac{1+a_{n-1}}{1-a_{n-1}-2a_{n-1}^2}$ or $a_n=\frac{a_{n-1}}{a_{n-1}+1}$ , can you explain me ? This is just a quadratic in $a_n$. Compute discriminant (it is $9a_{n-1}^2$) and use the classical formulas for roots of a quadratic and you get the two possibilies I gave. And since one is $>a_{n-1}$ and the other is $\le a_{n-1}$, we kept this second one.

I think we can solve it like this: (1)For $n=2$,I calculate $a_2=\frac{1}{3}$.During this course,I realized that maybe this equation could be factorised.So I get ${{2a_{n-1}-1}a_n+a_{n-1}{{a_{n-1}+1}a_n-a_{n-1}}=0$.And because $a_{n-1}\gt a_n$,so $$a_n=\frac{a_{n-1}}{a_{n-1}+1}$$.So$$ a_n=\frac{1}{n+1}$$(2)$$S_n=\sum_{i=1}^{n}\frac{1}{i+1}$$,so$$S_n\lt \int_{1}^{n+1}\\frac{1}{n+1}\dx\=ln {n+1}$$；$$S_n\gt\int_{2}^{n+2}\\frac{1}{n+2}=ln/frac{1}{n+2}$$

Chronokeeper wrote: I think we can solve it like this: (1)For $n=2$,I calculate $a_2=\frac{1}{3}$.During this course,I realized that maybe this equation could be factorised.So I get ${{2a_{n－1}－1}a_n+a_{n－1}{{a_{n－1}+1}a_n－a_{n－1})=0$.And because $a_{n－1}\gt a_n$,so $$a_n=\frac{a_{n-1}}{a_{n-1}+1}$$.So$$ a_n=\frac{1}{n+1}$$(2)$$S_n=\sum_{i=1}^{n}\frac{1}{i+1}$$,so$$S_n\lt \int_{1}^{n+1}\\frac{1}{n+1}\dx\=ln {n+1}$$；$$S_n\gt\int_{2}^{n+2}\\frac{1}{n+2}=ln/frac{1}{n+2}$$ If there any one would help me with the latex error?

Chronokeeper wrote: I think we can solve it like this: (1)For $n=2$,I calculate $a_2=\frac{1}{3}$.During this course,I realized that maybe this equation could be factorised.So I get $(2a_{n-1}-1)a_n+a_{n-1}((a_{n-1}+1)a_n-a_{n-1})=0$.And because $a_{n-1}\ge a_n$,so $$a_n=\frac{a_{n-1}}{a_{n-1}+1}$$.So$$ a_n=\frac{1}{n+1}$$(2)$$S_n=\sum_{i=1}^{n}\frac{1}{i+1}$$,so$$S_n\le \int_{1}^{n+1}\frac{1}{n+1}dx=ln {n+1}$$；$$S_n\ge\int_{2}^{n+2}\frac{1}{n+2}=ln\frac{1}{n+2}$$