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I'm quite stuck, truly. I have tried proving it directly, using induction, showing that $$n\ln (n) \leq \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big)$$ as well as using the fact that $$\sum ^n_{i=1}i\ln (i)\leq \sum ^n_{i=1}i\ln (n)=\frac{n(n-1)}{2}\ln (n )$$

but haven't been able to achieve much. The inequality does hold though, I've checked using desmos.

rtybase
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Sam
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  • What was your base case? $i=n-1, n=1,\ln 0$ isn't defined. – PinkyWay Mar 18 '20 at 22:48
  • @ms.VerkhovtsevaKatya If you are asking about the proof using induction, then I suppose the base case would be $\sum ^0{i=1}i\ln (i)=0\leq 0$. – Sam Mar 18 '20 at 22:51
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    @ms.VerkhovtsevaKatya What I've seen generally done is that the standard practice where the upper limit of a summation is less than the lower limit is that no terms are determined & added, with the result being treated as being $0$. This is similar to where you're dealing with a multiplication (e.g, $\pi{i=1}^{0}f(i)$), but with the result being treated as $1$ in that case. – John Omielan Mar 18 '20 at 22:52
  • @JohnOmelian, thank you for the response! – PinkyWay Mar 18 '20 at 22:54
  • You could apply Riemann sums, since $\int x \ln{x} dx = \frac{x^2}{4} (2 \ln{x} - 1) + C$ – rtybase Mar 18 '20 at 23:29

2 Answers2

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Look at the function $f(x)=x\ln{x}$ and recall Riemann sums (i.e. lower Riemann sum or lower Darboux sum). Given $f'(x)=\ln{x}+1$, $f(x)$ is monotone ascending for $x\geq1$. As a result $$\sum^{n-1}_{i=1} i\ln{i}= \sum^{n-1}_{i=1} i\ln{i} \left(i+1-i\right)=\\ \sum^{n-1}_{i=1} f(i) \left(i+1-i\right)\leq \int_{1}^{n}f(x)dx=\\ \frac{x^2}{4}(2\ln{x}-1)\Big|_{1}^n = \frac{n^2}{4}(2\ln{n}-1)+\frac{1}{4}=\\ \frac{n^2\ln{n}}{2}-\frac{n^2}{4}+\frac{1}{4}$$

rtybase
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It's not clear to me what the sticking point is. If you have proven that $$n\ln (n) \leq \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big)$$ then you are basically done, as this shows that the change in the left side of your desired inequality is less than the change of the right side.

To be explicit, let $f(n):=\sum_{i=1}^{n-1} i \ln (i)$ and let $g(n):= \frac{n^2}{2}\ln (n)-\frac{n^2}{4}-\frac{1}{4}$. Then $f(n+1)-f(n)$ is the left side of the above inequality, and $g(n+1)-g(n)$ is the right side. By induction, if $f(n)\leq g(n)$, then $$f(n+1) = \left(f(n+1)-f(n)\right)+f(n) \leq \left(g(n+1)-g(n)\right) + g(n) = g(n+1)$$ and together with $f(2)=0\leq 2\ln(2)-\frac{5}{4} = g(2)$ you're done. (The reason I chose $n=2$ as a base case rather than $n=1$ is because it makes the following argument work more smoothly. The $n=1$ case is trivial.)

The hard part is showing that the top inequality holds. Here's one way to do so: $$ \begin{align} & \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big) \\ = & \frac{n^2}{2}\left(\ln (n+1)-\ln(n)\right) + n\ln(n+1) + \frac{1}{2}\ln(n+1) - \frac{n}{2} - \frac{1}{4} \\ = & \frac{n^2}{2}\ln (1+\frac{1}{n}) + n\ln(n+1) + \frac{1}{2}\ln(n+1) - \frac{n}{2} - \frac{1}{4} \end{align} $$ Using $\ln(1+\frac{1}{n}) \geq \frac{1}{n}-\frac{1}{2n^2}$ (from the Taylor series expansion of $\ln$) on the first term, $\ln(n+1)\geq\ln(n)$ on the second term, and $\ln(n+1)\geq 1$ (for $n\geq 2$) for the third term, we find that the above is

$$\geq\frac{n}{2}-\frac{1}{4} + n\ln(n)+\frac{1}{2}-\frac{n}{2}-\frac{1}{4} = n\ln(n)$$as desired.

Yly
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