bump?...

Nominator $$\sum_{i=1}^{2n-1}\frac{1}{i(i+1)}=\sum_{i=1}^n\left(\frac{1}{i}-\frac{1}{i+1}\right)=1-\frac{1}{2n}=\frac{2n-1}{2n}$$Denominator $$\frac{1}{(n+1)n}+\frac{1}{(n+2)(2n-1)}=\frac{3n^2+4n-2}{n(n+1)(n+2)(2n-1)}=\frac{(n+1)(n+2)(2n-1)^2}{2(3n^2+4n-2)}$$For $n=999$ we have $3n^2+4n-2=2997997$ which is a prime number not dividing any of $(n+1),(n+2),(2n-1)^2$.

WolfusA wrote: Nominator $$\sum_{i=1}^{2n-1}\frac{1}{i(i+1)}=\sum_{i=1}^n\left(\frac{1}{i}-\frac{1}{i+1}\right)=1-\frac{1}{2n}=\frac{2n-1}{2n}$$Denominator $$\frac{1}{(n+1)n}+\frac{1}{(n+2)(2n-1)}=\frac{3n^2+4n-2}{n(n+1)(n+2)(2n-1)}=\frac{(n+1)(n+2)(2n-1)^2}{2(3n^2+4n-2)}$$For $n=999$ we have $3n^2+4n-2=2997997$ which is a prime number not dividing any of $(n+1),(n+2),(2n-1)^2$. You have also added 1/(2x3)+1/(4x5)+… Your statement is not correct.

CatalinBordea wrote: Show that $ \frac{\frac{1}{1\cdot 2} +\frac{1}{3\cdot 4}+\cdots +\frac1{1997\cdot 1998}}{\frac{2}{1000\cdot 1998} +\frac{1}{1001\cdot 1997}} $ is an integer number. Bogdan Enescu I am not sure if this is useful: $\sum\limits_{k=1}^{999}\frac{1}{(2k-1)\cdot 2k}=\sum\limits_{k=1}^{999}\left(\frac{1}{2k-1}-\frac 1{2k}\right)=\sum\limits_{k=1000}^{1998}\frac 1k=\frac 12\sum\limits_{k=1000}^{1998}\left(\frac 1k+\frac 1{2998-k}\right)=\frac 12\sum\limits_{k=1000}^{1998}\frac{2998}{k(2998-k)}$ $ \frac{\frac{1}{1\cdot 2} +\frac{1}{3\cdot 4}+\cdots +\frac1{1997\cdot 1998}}{\frac{2}{1000\cdot 1998} +\frac{1}{1001\cdot 1997}} =\frac 12\sum\limits_{k=1000}^{1998}\frac{2998}{k(2998-k)\left(\frac{2}{1000\cdot 1998} +\frac{1}{1001\cdot 1997}\right)}$

I do not understand the second above equality

Bump....

Is there something wrong???

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Sorry,I do not understand how to solve this ugly problem..

Emm...Notice that $1993$ is a prime. We need to prove $$\frac{2\cdot 1998!}{1000\cdot 1998}+\frac{1998!}{1001\cdot 1997}\mid \frac{1998!}{1\cdot 2} +\frac{1998!}{3\cdot 4}+\cdots +\frac{1998!}{1997\cdot 1998}$$However,$1993\mid \text{LHS}$ and $1993\nmid \text{RHS}$!!!

1993??? What is this?Please, explain.…

@above,I just choose a prime number $1993$ to prove the original number is not an integer.

PhilippineMonkey,if you can solve this anoying problem, please,explain in detail,how to solve... Thanks

Actually,I think this problem is wrong.Because I had proved $ \frac{\frac{1}{1\cdot 2} +\frac{1}{3\cdot 4}+\cdots +\frac1{1997\cdot 1998}}{\frac{2}{1000\cdot 1998} +\frac{1}{1001\cdot 1997}} $ isn't an integer.Maybe the post starter forgot something.

I understand that the denominație is 1/1000+1/1001+...+1/1998,but,how to solve,in finally?