1
$\begingroup$

In Knuth's "Concrete Mathematics" in chapter about numbers below equality is given $$H_n = \ln n + \gamma + \frac{1}{2n} - \frac{1}{12n^2} + \frac{\epsilon_n}{120n^4} $$ where $0 < \epsilon_n < 1$. For every $i$ holds $H_{i-1} < H_i$. Is this also true that for every $i$ holds: $$\frac{\epsilon_i}{120i^4} < \frac{\epsilon_{i-1}}{120(i-1)^4}$$. I was trying to prove that but I can't prove the dependency between $\epsilon_i$ and $\epsilon_{i-1}$. Where can I found a prove or a contrexample?

$\endgroup$
2
$\begingroup$

If you persue the approximation of $H_n$ a bit further you may find $$ H_n=\ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}+\frac1{240n^8}-\frac1{132n^{10}}+\mathcal O(\frac1{n^{12}}).$$ Hence for some constant $c$ (not explicitly specified in that Wikipedia page), we have $$ \epsilon_n=1-\frac1{252n^2}+\frac1{240n^4}-\frac1{132n^6}+\frac{\delta_n}{cn^8}$$ with $|\delta_n|<1$. Depending on the ecaxt value of $c$ this gives us that the sequence of the $\epsilon_n$ is strictly increasing either for all sufficiently large $n$ or possibly even for all $n$. Indeed, $e_{n+1}-\epsilon_n\approx\frac1{126n^3}$ already by the first term and the contribution of the last is at most $\approx \frac2{cn^8}$.

$\endgroup$
  • $\begingroup$ Thanks for your answer! But can you please specify what values $c$ and $\delta_n$ means? $\endgroup$ – Rop Jul 6 '14 at 11:42
  • $\begingroup$ The $\delta_n$ are like the $\epsilon_n$ in Knuth, the $c$ is like the $252$ in Knuth, but is here unspecified. But an explcit value can be found using the same techniques. In fact, to show only that $\epsilon_n$ is monotonic, it should suffice to make only the constant hidden in $\mathcal O(\frac1{n^6})$ explicit. $\endgroup$ – Hagen von Eitzen Jul 6 '14 at 11:47
  • $\begingroup$ Are you sure you didn't make any mistake in your post, because every time I try to evaluate that, I have a different result than yours. Maybe you can show more steps? $\endgroup$ – Rop Jul 6 '14 at 16:29
  • $\begingroup$ @rop, you need to learn to do some of the work yourself. All I can see is that the factor of 120 in $120 n^4$ was not carried along, it should go to the numerators that come after, so i get $$ \epsilon = 1 - \frac{10}{21 n^2} + \frac{1}{2 n^4} - \frac{10}{11 n^6} + \frac{\delta}{ n^8}. $$ If that is not quite right, you fix it. $\endgroup$ – Will Jagy Jul 6 '14 at 16:52
  • 1
    $\begingroup$ @WillJagy Now it seems to look like what I've get. But your not right. If I asked a question because I couldn't show something and the answer which is given has mistakes, then it doesn't make me any closer to the answer which I am looking for or even didn't give any knowledge, even if I will try to do something on my own, because it is just misleading. $\endgroup$ – Rop Jul 6 '14 at 17:04

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.