For the following equation, is there a better bound than the trivial bound I proposed $$ \Bigl( a \sum_{x=1}^n [\ x^d-(x-1)^d]\ \exp\Bigl\{-\frac{x^2}{\sigma^2}\Bigr\} +\sum_{x=1}^n x^{d-2}\exp\Bigl\{-\frac{x^2}{\sigma^2}\Bigr\} \Bigr) \\\leq (a+1)\sum_{x=1}^n x^d\exp\Bigl\{-\frac{x^2}{\sigma^2}\Bigr\} ~\text{as}~n~ \text{is large.} $$

This trivial upperbound seems large and I am looking for a tighter one. I thought about $ [\ x^d-(x-1)^d]\ = [\ (x-1+1)^d-(x-1)^d]$ but I don't have an idea how to continue!

I would be thankful for any help


1 Answer 1


Assuming $d\geq 1$, one has $$ x^d-(x-1)^d=x^d(1-(1-1/x)^d)\leq x^d(1-(1-d/x))=dx^{d-1} $$ using Bernoulli's inequality, which obtains a slightly better bound of $$ (ad+1)\sum_{x=1}^n x^{d-1}\exp(-x^2/\sigma^2). $$

  • $\begingroup$ then to continue as $$ (ad+1)\sum_{x=1}^n x^{d-1}\exp(-x^2/\sigma^2) \leq (ad+1)\sum_{x=1}^n x^{d}\exp(-x^2/\sigma^2) $$ or there is a better idea? $\endgroup$ Commented Jul 5, 2019 at 11:31
  • $\begingroup$ If you continue that way, you get an unconditionally worse bound than what you originally wrote. What I wrote is better when $\sigma^2$ isn't too small (as otherwise only the first few terms in the sum are nonnegligible) because $dx^{d-1}<x^d$ for $x>d$. So it really depends on how large $\sigma^2$ and $a$ are. $\endgroup$
    – J.G
    Commented Jul 5, 2019 at 17:30
  • $\begingroup$ In fact I am doing this is to lowerbound $\sigma^2$! here $x,d \rightarrow \infty$ but I can't say $x > d$! so,as I see, simplifying it will send me also to the same solution $\endgroup$ Commented Jul 5, 2019 at 18:06

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