2
$\begingroup$

Reading solutions to a practice exam, and I come across this: $$ O\left(\sum_{d \leq \sqrt{x}} {1 \over \sqrt{d}}\right) = O\left(x^{1/4}\right). $$ There are $O(\sqrt{x})$ terms in the sum, which are bounded between $x^{-1/4}$ and $1$. It looks like using the number of terms and the lower bound would give this estimate, but I don't see how that's valid.

$\endgroup$
  • 3
    $\begingroup$ Hint: Approximate the sum from above and below by a suitable integral (the function will be $\frac{1}{\sqrt{t}}$). $\endgroup$ – André Nicolas Dec 19 '13 at 23:32
3
$\begingroup$

By comparison to integrals, $$ 2\sqrt{n+1}-2=\int_1^{n+1}\frac{\mathrm{d}x}{\sqrt{x}}\le\sum_{k=1}^n\frac1{\sqrt{k}}\le\int_0^n\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{n} $$ So $$ \sum_{k=1}^n\frac1{\sqrt{k}}=O(\sqrt{n}) $$ plug in $n=\sqrt{d}$.

$\endgroup$
  • $\begingroup$ If I wanted to generalize this to other $p \in (0,1)$, the result is $\sum_{k=1}^n k^{-p} = O(n^{1-p})$, right? $\endgroup$ – A l'Maeaux Dec 19 '13 at 23:41
  • 2
    $\begingroup$ @Daniel: yes. The constant grows like $\frac1{1-p}$, so as $p\to1$, the constant gets very big. $\endgroup$ – robjohn Dec 19 '13 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.