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By playing with numbers, I discovered that

$n\ln(n) \approx \sum_{k=1}^n{\frac{n}{k}}$

And for $n = 1,2,...20,000$ the two quantities are almost neck-and-neck.

enter image description here It reminds me of the Stirling approximation. Does anyone know if this approximation has a name?

They may even converge... this is the ratio of the sum / log as a function of n: enter image description here

I forgot the calculus tests for divergence/convergence, so an example proving divergence would be great.

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    $\begingroup$ Note that $\sum_{i=k}^n n/k = n \sum_{i=1}^n 1/k$ and using the approximation $\sum_{k=1}^n f(k) \approx \int_1^n f(k) dk$ you get (more or less) your approximation. $\endgroup$
    – Zubzub
    Jan 18, 2017 at 8:22
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    $\begingroup$ $\lim_{m\to\infty}\left(\sum_1^m(1/k)-\log m\right)$ exists and is about $0.57721$ – it's called the Euler-Mascheroni constant, $\gamma$. $\endgroup$ Jan 18, 2017 at 8:26
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    $\begingroup$ By graphing $1/x$ you can convince yourself that : $$ \int_1^{n+1} 1/x\ dx\ \leq\ \sum_{i=1}^n 1/i\ \leq\ 1 + \int_1^n 1/x\ dx $$ Therefore : $$ \ln(n+1)\ \leq\ \sum_{i=1}^n 1/i\ \leq\ 1 + \ln(n) $$ $\endgroup$
    – Zubzub
    Jan 18, 2017 at 8:39

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Expanding my comments : Note that your claim $n\cdot \ln(n) \approx \sum_{i=1}^n \frac{n}{i}$ is equivalent to say $\ln(n) \approx \sum_{i=1}^n \frac{1}{i}$ simply by dividing both sides of the equality by $n$.

Now let us analyze the approximation $\ln(n) \approx \sum_{i=1}^n \frac{1}{i}$ :

By graphing $y = 1/x$ and noting that the summation can be thought as summing the areas of rectangles of base-length $1$ and height $1/x$ we can conclude that $$ \int_1^{n+1} \frac{1}{x}\ dx\ \leq\ \sum_{i=1}^n \frac{1}{i}\ \leq\ 1 + \int_1^n \frac{1}{x}\ dx $$ Computing the integral gives us : $$ \ln(n+1)\ \leq\ \sum_{i=1}^n \frac{1}{i}\ \leq\ 1 + \ln(n) $$

Now let's analyse the absolute difference between the two sides of the approximation. Namely : $ \sum_{i=1}^n \frac{1}{i} - \ln(n)$. From our inequalities we get : $$ \ln(n+1)- \ln(n) \ \leq\ \sum_{i=1}^n \frac{1}{i} -\ln(n)\ \leq\ 1 $$ Writing $\ln(n+1)- \ln(n) = \ln(\frac{n+1}{n}) = \ln(1+1/n)$ we get that : $$ \ln(1+ 1/n) \ \leq\ \sum_{i=1}^n \frac{1}{i} -\ln(n)\ \leq\ 1 $$ Applying $\lim_{n \rightarrow \infty}$ we see that the absolute difference is clearly bounded. (This doesn't prove the convergence but gives an intuition. Note that it acctually converges to $\gamma$) $$ 0 \ \leq\ \lim_{n \rightarrow \infty} \sum_{i=1}^n \left( \frac{1}{i} -\ln(n)\right) \ \leq\ 1 $$ Now analyzing the relative difference $\frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)}$, starting again from our inequalities : $$ \frac{\ln(n+1)}{\ln(n)}\ \leq\ \frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)}\ \leq\ \frac{1 + \ln(n)}{\ln(n)} $$ $$ \frac{\ln(n+1)}{\ln(n)}\ \leq\ \frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)}\ \leq\ 1+\frac{1}{\ln(n)} $$ Applying again $\lim_{n \rightarrow \infty}$ (and modulo some limits caclulations) we get $$ 1\leq\ \lim_{n \rightarrow \infty} \frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)} \leq 1 $$ and the sandwich theorem let us conclude that the relative difference goes indeed to $1$.

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