Let us define $H(k, n) = \displaystyle\sum_{i = 1}^{n} \frac{(\log i)^{k}}{i}$. We want to show that $H(k, n) - \displaystyle\frac{(\log n)^{k + 1}}{k + 1}$ converges as $n \rightarrow \infty$.

Notice that $\displaystyle\int_{1}^{n} \frac{(\log x)^{k}}{x} dx = \frac{(\log n)^{k + 1}}{k + 1}$. Therefore we wish to estimate the difference of the sum and integral. We can use Euler's summation formula to get:

$\displaystyle\sum_{i = 1}^{n} \frac{(\log i)^{k}}{i} - \displaystyle\int_{1}^{n} \frac{(\log x)^{k}}{x} dx = -(t - [t] - c)f(t)|^{n}_{1} + \int_{1}^{n} (t - [t] - c)t^{-2}(k - \log t)\log(t)^{k - 1} dt$

Choosing $c = 0$, the first term on the right side cancels. Since $t - [t] \leq 1$, and $\displaystyle\int_{1}^{n} t^{-2}(k - \log t)\log(t)^{k - 1} dt = \frac{\log(n)^{k}}{n}$, the integral on the right vanishes as $n \rightarrow \infty$ and the difference is... zero?

Obviously this is false but I can't see what's wrong with the argument. The formula for Euler's summation I found in Bateman's Analytic Number theory, page 47. Can anyone find the error here?

  • $\begingroup$ Where did you get $c$? $\endgroup$ – Norbert Oct 7 '12 at 23:35
  • $\begingroup$ If you check Bateman's book, it appears that it holds for any $c$. In an example he chooses $c = 0$ also. $\endgroup$ – Pedro Oct 7 '12 at 23:42

The integrand changes sign at $t=\mathrm e^k$. The fact that the positive and negative contributions cancel when you take out $t-[t]$ doesn't imply that they do if you don't. The sort of estimate you wanted to make would only be valid if the integrand is non-negative or if you derive the bound using its absolute value.

  • $\begingroup$ Hi @joriki may I ask you a conceptual question? which is not related to this question. $\endgroup$ – Seyhmus Güngören Oct 10 '12 at 13:30
  • $\begingroup$ @Seyhmus: Sure, but we shouldn't do that here in the comments. You can either invite me to a chat, or you can ask a regular question and then make me aware of it. $\endgroup$ – joriki Oct 10 '12 at 20:52
  • $\begingroup$ Ok clear. I was not much aware of inviting a person to chat. I will do in this way next time. Thanks. Meanwhile, I asked it as a question which you know, Lagrangian multipliers. $\endgroup$ – Seyhmus Güngören Oct 10 '12 at 20:55

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.