2
$\begingroup$

Evaluate the limit

$$ \lim_{n \to \infty} \frac {\sqrt{n+1}+\sqrt{n+2}+...+\sqrt{2n}}{n^{3/2}}. $$

Rearranging I can get

$$ \lim_{n \to \infty} \frac {\sqrt{\frac{n+1}{n}}+\sqrt{\frac{n+2}{n}}+...+\sqrt{\frac{2n}{n}}}{n}. $$

but I do not see how that helps me. Perhaps I use L'Hopital's rule?

$\endgroup$
  • 3
    $\begingroup$ Do you know the Riemann sum? $\endgroup$ – user63181 Jan 9 '15 at 21:54
7
$\begingroup$

$$\sum_{k=1}^n\frac{\sqrt{n+k}}{n^{3/2}}=\frac1n\sum_{k=1}^n\sqrt{1+\frac kn}\xrightarrow[n\to\infty]{}\int_0^1\sqrt{1+x}\;dx$$

If you like this answer please do not upvote it . Apparently someone in this site's administration believes some people(s) upvoted some of my posts unduely and some 695 (!) points from my reputation have been deleted, which seems to me exaggerated, but also there is at least once some 15 points ("accepted answer") that were deleted, as the number of points is not a multiple of ten. So someone who posted a question to which my answers were the accepted ones have their points given to me deleted.

I don't care about the points, and if you need to take them all off please do it, but leave alone trying to do some mathematics.

$\endgroup$
  • $\begingroup$ This is the perfect opportunity for $\sim$ $\endgroup$ – Mohamad Ali Baydoun Jan 9 '15 at 22:10
  • $\begingroup$ @MohammadAliBaydoun For what? $\endgroup$ – Timbuc Jan 9 '15 at 22:10
  • 1
    $\begingroup$ For $\sim$ rather than an arrow with $n \to \infty$. It's just really satisfying $\endgroup$ – Mohamad Ali Baydoun Jan 9 '15 at 22:10
  • $\begingroup$ @MohammadAliBaydoun Oh, I see. I really don't care for that a lot, and in this case the limit is exactly that integral, which is easily doable. $\endgroup$ – Timbuc Jan 9 '15 at 22:12
0
$\begingroup$

Using Stolz-Cesaro: $$\eqalign{ \lim_{n\to\infty}\frac{\sqrt{n+1}+\cdots+\sqrt{2n}}{\sqrt{n^3}} & = \lim_{n\to\infty}\frac{(\sqrt{(n+1)+1}+\cdots+\sqrt{2(n+1))}-(\sqrt{n+1}+\cdots+\sqrt{2n})}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr & = \lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cr &=\lim_{n\to\infty}\frac{\sqrt{2n+1}+\sqrt{2(n+1)}-\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{\sqrt{(n+1)^3}+\sqrt{n^3}}=\cdots }.$$

$\endgroup$

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.