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Calculate the limit of $$\lim_{n\to \infty} \sum_{k=1}^n \,\ln\left(\frac{k+n}{n}\right)^{\frac{n}{n^2+k^2}}$$ I believe that this thing converges but I couldn figured it out how to evaluate.

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    $\begingroup$ You can bring the exponent down and then think in terms of Riemann sums. $\endgroup$ – Anurag A May 28 at 8:15
  • $\begingroup$ I can see that but $$\frac{n}{n^2+k^2}$$ made me think to find boundaries $\endgroup$ – Sameen Shaw May 28 at 8:42
  • $\begingroup$ What does the exponent applies to ? $\endgroup$ – Yves Daoust May 28 at 12:24
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Rewrite $$S_n=\sum_{k=1}^n {\frac{n}{n^2+k^2}}\,\log\left(1+\frac{k}{n}\right)=\sum_{k=1}^n {\frac{1}{1+(\frac kn)^2}}\,\log\left(1+\frac{k}{n}\right)\frac1n$$ and then $$ \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n {\frac{1}{1+(\frac kn)^2}}\,\log\left(1+\frac{k}{n}\right)\frac1n=\int_0^1\frac{1}{1+x^2}\log(1+x)dx.$$ Under $x\to \tan x$, one has $$\int_0^1\frac{1}{1+x^2}\log(1+x)dx=\int_0^{\pi/4}\log(1+\tan x)dx=\frac\pi8\log2.$$

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  • $\begingroup$ I wanted to play with inequalities. Using the next level, I get $0.272196 < S_\infty < 0.272218$. $\endgroup$ – Claude Leibovici May 28 at 14:51
  • $\begingroup$ brilliant, thanks $\endgroup$ – Sameen Shaw May 31 at 9:47
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Rewriting $$S_n=\sum_{k=1}^n {\frac{n}{n^2+k^2}}\,\log\left(1+\frac{k}{n}\right)$$ and using as sharp bounds (have a look here)

$$\log\left(1+x\right) \leq \,\frac{x (x+6)}{2 (2 x+3)}$$ we have $$S_n <\sum_{k=1}^n \frac{k (k+6 n)}{2 (2 k+3 n) \left(k^2+n^2\right)}=\frac 1 {26}\sum_{k=1}^n \frac{20 k+9 n}{k^2+n^2}-\frac{27}{26}\sum_{k=1}^n \frac{1}{2 k+3 n}$$ $$\frac{20 k+9 n}{k^2+n^2}=\frac{20 k+9 n}{(k+i n)(k-in)}=\frac{\frac{9}{2}+10 i}{n+i k}+\frac{\frac{9}{2}-10 i}{n-i k}$$ make that we are ready to use genralized harmonic numbers $$\sum_{k=1}^n \frac{\frac{9}{2}+10 i}{n+i k}=\left(-10+\frac{9 i}{2}\right) \left(H_{-i n}-H_{(1-i) n}\right)$$ $$\sum_{k=1}^n \frac{\frac{9}{2}-10 i}{n-i k}=\left(-10-\frac{9 i}{2}\right) \left(H_{i n}-H_{(1+i) n}\right)$$ $$\sum_{k=1}^n \frac{1}{2 k+3 n}=\frac{1}{2} \left(H_{\frac{5 n}{2}}-H_{\frac{3 n}{2}}\right)$$

So, we have the bound. Now, using the asymptotics of the harmonic numbers, we have $$S_n <\frac{1}{104} \left(9 \pi -54 \log \left(\frac{5}{3}\right)+40 \log (2)\right)+\frac{7}{40 n}+O\left(\frac{1}{n^2}\right)$$ and the constant term is, numerically, $0.273227$.

We can do the same using (see here) $$\log\left(1+x\right) \geq \,\frac{3 x (x+2)}{x^2+6 x+6}$$

Repeating the process (I skip the steps) we end with $$S_n >\frac{3}{244} \left(7 \pi +64 \log (2)+32 \log (3)+4 \left(\sqrt{3}-8\right) \log (13)-4 \sqrt{3} \log \left(14-3 \sqrt{3}\right)\right)+\frac{9}{52 n}+O\left(\frac{1}{n^2}\right)$$ and the constant term is, numerically, $0.272092$.

So, to summarize $$\color{red}{0.272092 < S_\infty < 0.273227}$$

Computing exactly $S_{10^6}$ I got a value of $0.272198$.

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  • $\begingroup$ (+1) This is really cool. I've not see the bounds you used for the logarithm function. $\endgroup$ – Mark Viola May 29 at 13:26
  • $\begingroup$ @MarkViola. Using the next level of bounds, I get $0.272196<S∞<0.272218$. Amazing. In fact, I started working this problem because of you ! Cheers and take care. $\endgroup$ – Claude Leibovici May 29 at 13:44
  • $\begingroup$ Hi Claude! I am honored, but the bounds I typically have used are simple. These bounds transcend those. $\endgroup$ – Mark Viola May 29 at 13:46

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