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 September 1st, 2012, 10:15 PM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Divergence and Asymptotes 1. Prove or disprove : $\lim_{n \rightarrow \infty} \left | \sum_{k=2}^{n} \frac{1}{\ln(k)} - \text{Li}(n) \right |$ diverges. 2. Prove or disprove : $f(x)= \sum_{k=2}^{x} \frac{1}{\ln(k)} - \text{Li}(x)$ has an asymptote. Thank you for any help, Balarka .
 September 2nd, 2012, 01:17 AM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Divergence and Asymptotes HI mathbalarka: I assume $Li(n)$ is the polylogarithm?. The polylog is defined as $Li(n,x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k^{n}}$ But the problem simply states $Li(n)$. Perhaps I am misunderstanding, but which particular polylog is to be used?. In this case, it would be x=1 and mean $Li(n,1)=\sum_{k=1}^{\infty}\frac{1}{k^{n}}$. Then, we would have the limit $\lim_{n\to \infty}\sum_{k=1}^{\infty}\frac{1}{k^{n}}$ as part of the problem?. Another notation would be $Li_{n}(x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k^{n}}$ i.e. the dilogarithm: $Li_{2}(x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k^{2}}$ which, when x=1, would be the classic $Li_{2}(1)=\sum_{k=1}^{\infty}\frac{1}{n^{2}}=\frac {{\pi}^{2}}{6}$. In other words, we have to know which polylog we are dealing with. My apologies if I am not getting something obvious.
 September 2nd, 2012, 01:27 AM #3 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Divergence and Asymptotes No, $\text{Li}(x)$ is the offset logarithmic integral
 September 2nd, 2012, 01:30 AM #4 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Divergence and Asymptotes For 1, proving it disproving it diverges would probably use a technique similar to proving the Euler-Mascheroni constant converges. I don't know how to do that. It would be somehow showing that $\int_{2}^{\infty}$$\frac{1}{\lfloor\log(x)\rfloor}-\frac{1}{log(x)}$$dx$ diverges. The Euler-Mascheroni constant has x in place of log(x). The differences probably do not decrease fast enough as it's logarithmic.

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