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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?

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

1 Answer 1

up vote 2 down vote accepted
+100

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.

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Hi @joriki may I ask you a conceptual question? which is not related to this question. –  Seyhmus Güngören Oct 10 '12 at 13:30
    
@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. –  joriki Oct 10 '12 at 20:52
    
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. –  Seyhmus Güngören Oct 10 '12 at 20:55

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