Tag Info

New answers tagged

1

You can see the sum as the integral of the step function $$e^{-(\lfloor x\rfloor-c)^2/b}.$$ In the increasing area ($x<c$), $$e^{-(x-1-c)^2/b}\le e^{-(\lfloor x\rfloor-c)^2/b}\le e^{-(x-c)^2/b},$$ and conversely ($x>c$), $$e^{-(x-c)^2/b}\le e^{-(\lfloor x\rfloor-c)^2/b}\le e^{-(x-1-c)^2/b}.$$ Integrating these bracketings with the relevant bounds, ...


1

A possible upper bound may be obtained by considering geometric sums as follows $$\sum_{n=1}^{N}e^{-(n-c)^{2}}= \sum_{n=1}^{N}e^{2cn-c^{2}-n^{2}}\leq \sum_{n=1}^{N}e^{2cn-c^{2}}=e^{-c^{2}}\sum_{n=1}^{N}\left( e^{2c}\right) ^{n}=e^{-c^{2}}e^{2c}\frac{1-e^{N(2c)}}{1-e^{2c}}(\leq e^{2c}\frac{1-e^{2Nc}% }{1-e^{2c}})$$ The last inequality is optional (it depends ...


1

Notice that each term in the sum is bounded by $$ e^{(\sum_{i}x_i)^2} = e^{\sum_{i} x_i^2 + \sum_{i \neq j} x_i x_j} \leq e^{\epsilon + \sum_{i \neq j} x_i x_j} $$ Then observe that the vector formed by the $x_i$ lies inside a ball of radius $\epsilon$, so each $x_i < \sqrt{\epsilon}$ and we can bound the sum $$ \sum_{i \neq j} x_i x_j < \sum_{i \neq ...


1

Hint: Use $$\frac{n!}{(n-k)!} = n(n-1)\dots(n-k+1) = n^k(1+o(1))\approx n^k$$ and $$ \big(1-\frac{\lambda}{n}\big)^k = 1+o(1) \approx 1. $$



Top 50 recent answers are included