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Show that the sequence $$x_n=\sum_{k=1}^n\frac{\ln k}{k}-\frac{(\ln n)^2}{2}$$ is convergent. And the given hint is to show convergence of the series $$\sum_{p=1}^\infty(x_{p+1}-x_p).$$ I got into trouble when trying to prove $(x_{p+1}-x_p)=o\left(\dfrac{1}{p^{3/2}}\right)$. So I wonder if there exist a simple way to get the convergence of the complicated series $$\sum_{p=1}^\infty(x_{p+1}-x_p)=\sum_{p=1}^\infty\left[\frac{\ln (p+1)}{p+1}-\frac{(\ln(p+1))^2}{2}+\frac{(\ln p)^2}{2}\right].$$

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    $\begingroup$ Can you use the mean value theorem? (If so, applying it to $\frac{(\ln b)^2}{2} - \frac{(\ln a)^2}{2}$ is a good start.) $\endgroup$ – Daniel Fischer Nov 29 '17 at 16:59
  • $\begingroup$ Can you use the integral test? $\endgroup$ – Jacky Chong Nov 29 '17 at 17:05
  • $\begingroup$ @DanielFischer Thanks for your hint! That's an elegant solution. $\endgroup$ – username Nov 30 '17 at 3:01
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You might be interested in a different approach: The expression equals

$$\frac{\ln 2}{2} + \sum_{k=3}^{n-1} \frac{\ln k}{k} +\frac{\ln n}{n} -\frac{(\ln n)^2}{2}.$$

Why did I break the sum up that way? Hopefully the motivation will be clear below.

Because $\dfrac{\ln n}{n}\to 0,$ we can neglect it. Next note

$$\frac{(\ln n)^2}{2} = \int_1^n \frac{\ln x}{x}\, dx = \int_1^3 \frac{\ln x}{x}\, dx + \int_3^n \frac{\ln x}{x}\, dx.$$

It follows that we only need to show that the limit of the following expression exists:

$$\tag 1 \sum_{k=3}^{n-1} \frac{\ln k}{k} - \int_3^n \frac{\ln x}{x}\, dx.$$

I focus on the number $3$ because $(\ln x)/x$ is decreasing on $[e,\infty).$ So we are in the familiar situation with upper sums vs integrals in the case of a positive decreasing integrand. It follows that $(1)$ increases with $n$ and is bounded above by $(\ln 3)/3.$ Hence the limit of $(1)$ exists and we're done.

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By summation by parts, for any $n>1$ we have $$\begin{eqnarray*} L_n = \sum_{k=1}^{n}\frac{\log k}{k} &=& H_n \log(n) -\sum_{k=1}^{n-1}H_k \log\left(1+\frac{1}{k}\right)\\&=&\log(n)^2+\gamma\log(n)+O\left(\frac{\log n}{n}\right)-\sum_{k=1}^{n-1}\frac{H_k}{k}+\sum_{k=1}^{n-1}H_k\left[\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right] \\&=&\log(n)^2+O\left(1\right)-\sum_{k=1}^{n-1}\frac{\log k}{k}+\sum_{k=1}^{n-1}H_k\left[\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right]\end{eqnarray*}\tag{A}$$ hence $2\,L_n=\log(n)^2+C+o(1)$, since the series $K=\sum_{k\geq 1}H_k\left[\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right]$ is absolutely convergent. An alternative (similar) approach is to exploit $$ \log(n)= H_n -\gamma-\frac{1}{2n}+O\left(\frac{1}{12n^2}\right) $$ $$ \sum_{n=1}^{N}\frac{H_n}{n}=\frac{H_n^2+H_n^{(2)}}{2} $$ $$ \lim_{N\to +\infty}\left[-\frac{\log^2 N}{2}+\sum_{n=1}^{N}\frac{H_n-\gamma-\frac{1}{2n}}{n}\right]=-\frac{\gamma^2}{2}.$$

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