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Jul
26
accepted Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?
Jul
7
comment Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?
I had the following idea for a lower bound, do you thing it carries on? $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}=\sum\limits_{k=0}^n \frac{k+1-k}{k+1}\left\{ n \atop k\right\}=B_n-\sum\limits_{k=0}^n \frac{k}{k+1}\left\{ n \atop k\right\}$. But $\sum\limits_{k=0}^n \frac{k}{k+1}\left\{ n \atop k\right\}<B_{n+1}-B_n$, so $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\} >2B_n-B_{n+1}$.
Jul
7
comment Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?
@leonbloy exact would be perfect, but asymtotical would be fine as well.
Jul
7
comment Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?
@J.M. thank you! I edited the post
Jul
7
revised Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?
deleted 60 characters in body
Jul
7
asked Expression for $\sum\limits_{k=0}^n \frac1{k+1}\left\{ n \atop k\right\}$?
May
2
accepted Prove or give a counter-example for the inequality
May
2
comment Prove or give a counter-example for the inequality
you are right. By mistake I was working with >f := 2*((1+(1+x*(n-1))*(1+x*(s-1)))^(1/2)-1-x*(s-1))/x in maple (and in paper) which has an extra 1 in the sqrt that introduces the extra -2 in the differentiation. Again you are very right! Thanks!
May
2
comment Prove or give a counter-example for the inequality
Is there a possibility that in the numerator of the expression of $f^'$ you provided, instead of -2 we should have -4? I just wanted to verify the calculations again (using both maple and paper) and it seems this way... This -4 might make some difference in applying L'Hospital (have not checked this yet)...
Apr
30
comment Prove or give a counter-example for the inequality
Yes you are correct, the monotonicity will solve the problem. So the solution is complete.
Apr
30
comment Prove or give a counter-example for the inequality
Yes, this is correct as $\gamma$ approaches $0$ from the right. But it is the half way there. We have to check also what happens as $\gamma$ tends to 1.
Apr
30
comment Prove or give a counter-example for the inequality
Say now that you would like to see what happens when $\gamma$ approaches $0$, you will then have a situation in which $\frac{2}{0}0$ in which you can not say anything.
Apr
30
comment Prove or give a counter-example for the inequality
Say you try to bound the left hand side from above: $\frac{2}{\gamma}[\sqrt{(1+\gamma (n-1))(1+\gamma (s -1))}-(1+\gamma (s -1))] \leq \frac{2}{\gamma}[\sqrt{(1+\gamma (n-1))^2}-(1+\gamma (s -1))] \leq \frac{2}{\gamma}\gamma(n-s)$ which is not a good bound
Apr
30
comment Prove or give a counter-example for the inequality
@Peter Tamaroff: you can use theory of convex functions if you wish.
Apr
30
comment Prove or give a counter-example for the inequality
@Peter Tamaroff: Any correct proof or counter-example provided will do.
Apr
30
asked Prove or give a counter-example for the inequality
Apr
10
comment Find the optima of the sum
yes you have a point here, $n$ and $s$ are fixed so be $n-s$.
Apr
10
asked Find the optima of the sum
May
5
accepted sums of reciprocals
May
5
comment sums of reciprocals
yes the counter-example works!