1
$\begingroup$

Let $n_1,\ldots,n_{m+1}$ be natural numbers, such that their sum is equal to some constant $C$.

I would like to calculate $$ 1+\frac{1}{\sqrt{n_ 1}}\left( 1+ \frac{1}{\sqrt{n_2}}+\frac{1}{n_2\sqrt n_3}+\cdots+\frac{1}{n_2\cdots n_m\sqrt{n_{m+1}}} \right) $$

  • 1
    This isn't typesetting on my machine. Seems to be in need of an edit.2012-04-18
  • 0
    Don't you mean "$1+\frac{1}{\sqrt{n_ 1}}\left(1+ \frac{1}{n_2 \sqrt{n_2}}+\cdots+\frac{1}{n_2\cdots n_{\mathbf{m+1}}\sqrt{n_{m+1}} }\right)$?" That is, isn't the product in the denominator of the last term "$n_2\cdots n_{\mathbf{m+1}}\sqrt{n_{m+1}}$?"2012-04-18
  • 0
    Oh thank you. It was typo.2012-04-18
  • 0
    Do you have some reason to believe that quantity really depends only on $C$? Have you done any numerical experiments to test this? For example, does $n_1=1$, $n_2=4$, $n_3=9$ give the same answer as $n_1=4$, $n_2=9$, $n_3=1$? In both cases, $C=14$.2012-04-18

0 Answers 0