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Let me first state the question and then provide some context and my approach. I'm trying to evaluate limit of the two following sequences : $$ S_n = \sum_{1\leq k \leq n}\frac{2k^2-3kn-3n^2}{(k+3n)(k^2-2kn+5n^2)}, \quad P_n=n\cdot\sum_{1\leq k \leq n}\frac{n+k}{(k-2n)(k^2+kn+n^2)} $$

Ok now, for context, I had the first sum to evaluate on an exam about 2-3 years ago while I was an undergrad, the second sequence was on the exam the year before I took the class. At the time, when I saw the second sum while working to prepare the exam, I couldn't solve it and thought to myself, "if this shows up on the exam, I most definitely will skip that". A few weeks later the (quite similar) first sum did show up on the exam, and I did skip it. At the time I tried really hard but I eventually gave up. This brings us back to today, I'm still studying math, albeit at a higher level now, and I somehow thought about this problem. I've learned a lot since then, yet I still have no idea how to compute those two limits. It should be a fairly easy problem, I mean the class was Real analysis 2 or 3 and given during my second year of university, so really nothing fancy.

My approach back then, which I replicated today, to no avail, was manipulating these sums to find that they were Riemann sums of some well-known or easy integral. I did not succeed in doing so. Another thing I attempted was partial fraction decomposition but either I did it wrong every single time, or it doesn't help here. I just don't know what to do anymore. I suspect solving one of the two would give enough insight to be able to solve the other because they're very similar. My intuition tells me if they are indeed Riemann sums, the function to integrate must be the same, maybe scaled or over a different interval or something but they look related, i.e. they're of the form $\sum_{1\leq k\leq n}\frac{ak^2+bkn+cn^2}{(dk+en)(fk^2+gkn+hn^2)}$ for $a,b,c,d,e,f,g,h\in \mathbb{R}$.

t_kln
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2 Answers2

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Note that

$$S_n = \sum_{1\leq k \leq n}\frac{2k^2-3kn-3n^2}{(k+3n)(k^2-2kn+5n^2)}=\frac1n\sum_{1\leq k \leq n}\frac{2\left(\frac k n\right)^2-3\frac kn-3}{\left(\frac kn+3\right)\left(\left(\frac k n\right)^2-2\frac kn+5\right)}$$

and

$$P_n=n\cdot\sum_{1\leq k \leq n}\frac{n+k}{(k-2n)(k^2+kn+n^2)} =\frac1n\sum_{1\leq k \leq n}\frac{1+\frac kn}{\left(\frac k n-2\right)\left(\left(\frac k n\right)^2+\frac k n+1\right)}$$

and we can use Riemann sum.

Refer also to:

user
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  • Damn... It was "that easy".... I was never good at computation, but I got to say, this one hurts. – t_kln Apr 17 '23 at 22:55
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    @notananalyst Yes these cases are indeed simple, we just need to divide for $n$ or $n^2$ in order to get the $k/n$ terms or constant terms. – user Apr 17 '23 at 22:58
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Suppose that you want to compute the partial sums $$S_n = \sum_{k=1}^n\frac{2k^2-3kn-3n^2}{(k+3n)(k^2-2kn+5n^2)}$$ Write $$2k^2-3kn-3n^2=2(k-a n)(k-b n)$$ $$k^2-2kn+5n^2=(k-cn)(b-dn)$$ whatever the roots of the quadratic equations could be (real or complex).

Now, using partial fraction decomposition $$\frac{2k^2-3kn-3n^2}{(k+3n)(k^2-2kn+5n^2)}=\frac{A}{k+3n}+\frac{B}{k-cn}+\frac{C}{k-d n}$$ So, three terms looking like $$\sum_{k=1}^n \frac 1{k+fn}=H_{(f+1) n}-H_{f n}$$

The values are $$a=\frac{3-\sqrt{33}}{4} \qquad b=\frac{3+\sqrt{33}}{4} \qquad c=1-2 i \qquad d=1+2i$$ $$A=\frac{6}{5}\qquad\qquad B=\frac{2}{5}-\frac{11 }{20}i\qquad\qquad C=\frac{2}{5}+\frac{11 }{20}i$$

Using the asymptotics of the harmonic numbers for large values of $n$, it will give $$S_n=\frac{1}{20} \left(8 \log \left(\frac{256}{135}\right)+22 \tan ^{-1}(2)-11\pi\right)-\frac{1}{40 n}+\frac{203}{7200 n^2}+\frac{196717}{259200000 n^4}+O\left(\frac{1}{n^6}\right)$$

Claude Leibovici
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