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The problem is to find the limit of:

$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$

A the series is finite, it looks as if it would be required to find the sum of the series - however, I have to find the limit. It resembles for me:

$$\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n^2-kn }{(k+n)^3-2kn^2-2k^2n}=\\ \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)^3-2kn(n+k)}=\\ \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)((k+n)^2-2kn)}$$

but I don't know what to do next and how to solve it. I would appreciate your help.

Edit: Does it has something in common with Riemann sum?

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  • $\begingroup$ I don't know if it could help (I am stuck) but $ \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}=\frac{n}{k^2+n^2}+\frac{1}{k+n}$ $\endgroup$ Oct 12, 2020 at 7:05
  • $\begingroup$ Is it possible that the upper end of $\lfloor n + \sqrt{n}\rfloor$ is a red herring, and you are supposed to prove that the series is divergent through a bizarre comparison to the harmonic series? $\endgroup$ Oct 12, 2020 at 7:11
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    $\begingroup$ @user2661923. First time I hear about a red herring ! Amazing. $\endgroup$ Oct 12, 2020 at 7:17
  • $\begingroup$ @ClaudeLeibovici Your comment is ironic since one of the clues was your rating. If 168k+ rep is having trouble, something may be up. $\endgroup$ Oct 12, 2020 at 7:18
  • $\begingroup$ user2661923, I am not sure whether it is going to end up this way. It would be more cautious to consider other options first. What about Claude Leibovici modification and the sum of Rieman integrals from 1 to $n+\sqrt(n)$? However, we have the floor here... $\endgroup$
    – Funny
    Oct 12, 2020 at 7:22

3 Answers 3

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Attempts

Using the integral test, we can show the convergence since $$I=\int \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}\,dk=\int \left(\frac{n}{k^2+n^2}+\frac{1}{k+n}\right)\,dk=$$ $$I=\log (k+n)-\tan ^{-1}\left(\frac{n}{k}\right)$$

Integrating between $k=1$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have $$\tan ^{-1}\left(\frac{n \left(\left\lfloor n+\sqrt{n}\right\rfloor -1\right)}{\left\lfloor n+\sqrt{n}\right\rfloor +n^2}\right)+\log \left(\frac{\left\lfloor n+\sqrt{n}\right\rfloor +n}{n+1}\right)$$ which is asymptotic to $\frac \pi 4+\log(2)\approx 1.47855$.

Integrating between $k=0$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have $$\log \left(\frac{\left\lfloor n+\sqrt{n}\right\rfloor +n}{n}\right)-\tan ^{-1}\left(\frac{n}{\left\lfloor n+\sqrt{n}\right\rfloor }\right)+\frac{\pi }{2}$$ which shows the same asymptotic value.

On the other hand, we can write the summation as $$-\frac{1}{2} i H_{\left\lfloor n+\sqrt{n}\right\rfloor -i n}+\frac{1}{2} i H_{i n+\left\lfloor n+\sqrt{n}\right\rfloor }+H_{n+\left\lfloor n+\sqrt{n}\right\rfloor }-H_n-\frac{1}{2 n}+\frac{1}{2} \pi \coth (\pi n)$$ and using the asymptotics of harmonic numbers, the same result.

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First, we can omit the $\sqrt{n}$ part and only sum up to $n$. To see this, $$ \sum_{k=n}^{n+\sqrt{n}} x_n \leq \sum_{k=n}^{n+\sqrt{n}} \frac{4(n+\sqrt{n})^2}{4n^3} \leq \sum_{k=n}^{n+\sqrt{n}} \frac{(n+n)^2}{n^3} \leq \sum_{k=n}^{n+\sqrt{n}} \frac{4}{n} \leq 4\frac{\sqrt{n}}{n} \mapsto 0 $$ The rest is the Riemannian sum of the integral $$ \int_0^1 \frac{x^2 + x + 2}{x^3 + x^2 + x + 1} \mathrm{d}x = \int_0^1 \frac{1}{1+x} \mathrm{d}x + \int_0^1 \frac{1}{1+x^2}\mathrm{d}x. $$ This can be easily computed using calculus.

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    $\begingroup$ I found the same result adding $10^9$ terms of the original series. By the way, the integrand can be written as $1/(1 + x) + 1/(1 + x^2)$ which lead immediately to the result. $\endgroup$
    – Raffaele
    Oct 12, 2020 at 8:55
  • $\begingroup$ Thanks for the decomposition, I added it into the answer, hope it's ok. $\endgroup$ Oct 12, 2020 at 9:07
  • $\begingroup$ Thank you very much, it looks very nice for me. However, why the interval of integration is (0,1)? It is a very important question to me, as I have not found the answer to this yet in other examples. Additionally, could you explain more clearly, how do you omit $\sqrt{n}$ $\endgroup$
    – Funny
    Oct 12, 2020 at 11:51
  • $\begingroup$ @Funny I could explain all in great detail, but would it not be better if you take this as some kind of hints, and complete it? Why from 0 to 1... Well, the expression can be rewritten to $\sum_1^n \frac{1}{n} \frac{(k/n)^2 + (k/n) + 2}{(k/n)^3 + (k/n)^2 + (k/n) + 1}$. If $x=k/n$, what is the range for $x$? $\endgroup$ Oct 12, 2020 at 12:25
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    $\begingroup$ Looking at the growth of the nominator: from 0 to 1. Thank you! $\endgroup$
    – Funny
    Oct 12, 2020 at 12:31
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Thanks to Peter Franek I would try to calculate the limit.

The sum can be rewritten as: $$\ \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3} = \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) $$

$$\ a_n=\frac{n}{k^2+n^2} = \frac{n}{n^2}\cdot\frac{1}{\frac{k^2}{n^2}+1}$$

$$\ b_n=\frac{1}{k+n} = \frac{1}{n}\cdot\frac{1}{\frac{k}{n}+1} $$

Let us consider $f(x) = \frac{1}{x^2+1} $ and $g(x)=\frac{1}{x+1}$

$$\lim_{n\to\infty}a_n=\int_0^1f(x)dx=\int_0^1\frac{1}{x^2+1}dx=\arctan(x)|_0^1=\frac{\pi}{4} \\\lim_{n\to\infty}b_n=\int_0^1g(x)dx=\int_0^1\frac{1}{x+1}dx = \int_0^1\frac{c'(x)}{c(x)}=\ln(x+1)|_0^1=\ln(2)$$

As the element $\sqrt(n)$ becomes unsignificant for $n\to\infty$ as $\frac{\sqrt(n)}{n}\to0$, the limit is:

$$ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) = \frac{\pi}{4}+\ln(2)$$

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  • $\begingroup$ Yes, just remove the $-1$. Also, instead of $\lim a_n$, you probably want to write $\lim \sum_1^n a_n$. And what "becomes insignificant", is the sum $\sum_n^{n+\sqrt{n}}$. And missing $\lim$ in the last equation. $\endgroup$ Oct 13, 2020 at 5:59
  • $\begingroup$ Thank you @PeterFranek your corrections. However, I have some questions about your notes. Firstly, I am quite confuesd as you insist on \sum_1^na_n, because as far as I'm corncerned, the Rieman sum approximate the limit of the sequence and I haven't encountered the sum notation. Is there a difference. Secondly, the insignificance is the result of the value of the sum \sum-n^{n+\sqrt{n}}, that doesn't change the result, isn't it? $\endgroup$
    – Funny
    Oct 13, 2020 at 10:18
  • $\begingroup$ Ok, I edited my answer to add details about the $\sqrt{n}$. The rest about Riemannian sum, please read about it here, for instance en.wikipedia.org/wiki/Riemann_sum The sum is a way how to approximate (or define) definite integrals. $\endgroup$ Oct 13, 2020 at 10:28

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