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$$\lim_{n \to \infty}\frac{1}{\log n}\sum_{k=1}^{n^2}1/k$$

My attempt: $$\lim_{n \to \infty}\sum_{k=1}^{n^2}1/k=\lim_{n \to\infty} \int_1^{n^2}1/xdx$$

Then this converges to infinity as it is harmonic series, and $$\lim_{n \to \infty}\log n \to \infty $$ Using L'hopital Rule, $$\lim_{n \to \infty}\frac{1}{\log n}\sum_{k=1}^{n^2}1/k=\lim_{n \to\infty}\frac{1/n^2-1}{1/n}=1$$

My question: Is this correct? If so is it possible everytime to make such series of decreasing sequence to integration ( integral test) whose endpoints are this way. If not correct please help me.

dhashtp
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    What do you think $\int\limits_1^{n^2}\frac1x,dx$ is? – Henry May 08 '24 at 08:35
  • I do not understand the sentence "is it possible everytime to make such series of decreasing sequence to integration ( integral test) whose endpoints are this way". Anyway, the numerator $1/n^2-1$ in your use of L'Hôpital's rule is (at best!) unclear. $t\mapsto\sum_{k=1}^{t^2}1/k$, whatever that means for a real variable $t$, is not differentiable. – Anne Bauval May 08 '24 at 08:38
  • @Henry it is $\log n^2$ – dhashtp May 08 '24 at 08:47
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    and what is $\frac{\log (n^2)}{\log(n)}$? – Henry May 08 '24 at 08:49
  • AnneBauval in numerator I used derivative of integration is the function itself $\Sigma_{k=1}^{t^2}1/k=\int_1^{n^2}1/xdx$ derivative of this I used as $1/n^2-1$. I think this is mistake please point out. If I am wrong – dhashtp May 08 '24 at 08:51
  • Henry Thanks for clarifying, my mistake was I got confused with limit and integration. So it is $\frac{\log n^2}{\log n}=2$ hence limit is 2. – dhashtp May 08 '24 at 08:53
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    $\sum_{k=1}^{n^2}1/k\ne\int_1^{n^2}1/xdx$. Your mistake was here. – Anne Bauval May 08 '24 at 08:54
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    @AnneBauval Thanks. That is what exactly I was asking whether this integration and sum straight away can be written like this. I got my mistake. It is not true. – dhashtp May 08 '24 at 08:56
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    Indeed the limit is $2$. For $n=1000$ you get about $2.08$ – Henry May 08 '24 at 08:56

2 Answers2

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There is no problem with your $$\lim_{n \to \infty}\sum_{k=1}^{n^2}1/k=\lim_{n \to\infty} \int_1^{n^2}1/xdx$$ (by the integral test, since the RHS $\int_1^\infty1/xdx$ equals $\infty$, so does the LHS $\sum_{k=1}^\infty1/k$).

Your only mistake is in removing the limits in this equality (which, if correct, would then perfectly justify your use of L'Hôpital's rule). $$\sum_{k=1}^{n^2}1/k\ne\int_1^{n^2}1/xdx.$$

For a correct calculation of your limit, use the asymptotic estimate of the harmonic number: as $n\to\infty$, $$H_n:=\sum_{k=1}^n\frac1k\sim\ln n$$ hence $$\frac1{\ln n}H_{n^2}\sim\frac1{\ln n}\ln(n^2)=2.$$

Anne Bauval
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As an alternative, by Stoltz-Cesaro

$$\frac{\sum_{k=1}^{(n+1)^2}1/k-\sum_{k=1}^{n^2}1/k}{\log (n+1)-\log n}=\frac{\frac1{n^2+1}+\ldots+\frac1{(n+1)^2}}{\log\left(1+\frac1n\right)}=2\frac{\frac{\frac{n^2}{n^2+1}+\ldots+\frac{n^2}{(n+1)^2}}{2n}}{\log\left(1+\frac1n\right)^n} \to 2$$

user
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  • Your answer is essentially a duplicate of that one. I just inserted a link to the latter in my CW answer. – Anne Bauval May 08 '24 at 09:28
  • @AnneBauval Yes thanks, this is essentially the same idea of course, here a little tricky for the $n^2$ term which requires some more effort and idea. – user May 08 '24 at 09:31
  • Indeed. Using the method of the linked post seems more natural, and in any case less tricky. Another point is that your answer does not really adress the OP's question (if it did, the question could be closed as a duplicate). The OP's main question was not about how to find the limit, but whether their use of L'Hopital was correct. – Anne Bauval May 08 '24 at 09:44
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    @AnneBauval Yes indeed the mistake was alredy pointed out in comments and other answer. Anyway I suppose that also the alternative way can be instructive and important for the asker. What do you mean by "using the method of the linked post"? Isn't it the application of teh same theorem but for another case? Thanks – user May 08 '24 at 09:52
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    I agree with everything in your last comment. – Anne Bauval May 08 '24 at 09:54
  • @AnneBauval Well done! Thanks again. Regards – user May 08 '24 at 10:06