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?


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 $\approx \frac2{cn^8}$.

  • $\begingroup$ 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
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    $\begingroup$ @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

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