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I want to compute this series, $$\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}.$$ I don't know how to contiune after factoring, which yields $\frac{1}{k(k-1)^2(k+1)^2(k-2)^2(k+2)^2}.$

I know that $\frac{1}{k(k-1)}$ telescopes. However, I can't apply this here, as the product of the sum is not equal to the sum of the product. I only want hints for now.

Thomas Andrews
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dude076
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3 Answers3

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Well, you seem to know how to telescope summing: $$\frac1{k(k-1)} = \frac1{k-1} - \frac1k$$

Similarly, note $$\frac1{(k-2)^2(k+2)^2} = \frac1{8k}\left(\frac1{(k-2)^2}- \frac1{(k+2)^2}\right)$$

$$\implies \frac1{(k-2)^2(k-1)^2k(k+1)^2(k+2)^2} = \frac1{8(k-1)^2k^2(k+1)^2}\left(\frac1{(k-2)^2}- \frac1{(k+2)^2}\right)$$

which can be noted to telescope, hence the sum is $\dfrac1{8\cdot 2^2\cdot3^2\cdot4^2}\cdot\dfrac1{1^2}=\dfrac1{4608}$.

Macavity
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Perform a partial fraction decomposition; specifically, find constants $A_1, A_2, \ldots, A_9$ such that $$\begin{align}\frac{1}{k(k^4-5k^2+4)^2} &= \frac{A_1}{k} + \frac{A_2}{k-1} + \frac{A_3}{k+1} + \frac{A_4}{k-2} + \frac{A_5}{k+2} \\&\quad+ \frac{A_6}{(k-1)^2} + \frac{A_7}{(k+1)^2} + \frac{A_8}{(k-2)^2} + \frac{A_9}{(k+2)^2}. \end{align}$$

heropup
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  • OP, @JeanMarie I am not sure if partial fractions alone helps - we cant really sum up the harmonic series terms above just because we have the coefficients solved. Unless the coefficients magically vanish /cancel, or we can club terms to do telescoping. – Macavity Apr 16 '21 at 07:24
  • @Macavity You are right : indeed there is a problem with harmonic series. – Jean Marie Apr 16 '21 at 07:33
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Let us work on the partial sum (with a fixed "large" $n$);

$$S_n:=\sum _{k=3}^{n}\:\frac{1}{k\left(k^4-5k^2+4\right)^2}=\sum _{k=3}^{n}\frac{1}{k(k-1)^2(k+1)^2(k-2)^2(k+2)^2}\tag{1}$$

Here is a way to use the partial fraction decomposition:

$$\dfrac{1}{k (k^4 - 5 k^2 + 4)^2} =$$

$$= \dfrac{1}{16 k} -\dfrac{1}{54 (k - 1)} - \dfrac{1}{54 (k + 1)} - \dfrac{11}{864 (k - 2)}-\dfrac{11}{864 (k + 2)} + \dfrac{1}{36 (k- 1)^2} - \dfrac{1}{36 (k+ 1)^2} - \dfrac{1}{288 (k + 2)^2} + \dfrac{1}{288 (k - 2)^2}\tag{2}$$

(I obtained (2) by the following Wolfram Alpha request Partial Fraction 1/(k(k^4-5*k^2+4)^2))

One can organize the summation (1) using two classical results involving transcendental numbers $\gamma$ and $\pi$: see here and here).

These results can be used by taking into account the fact that most of all summations don't begin at their first term, needing therefore a compensation:

$$\begin{cases} \approx +\dfrac{1}{16}(\ln(n)+\gamma)-\dfrac{1}{16}(1+\tfrac12)\\ \approx -\dfrac{1}{54}(\ln(n-1)+\gamma)+\dfrac{1}{54}\\ \approx -\dfrac{1}{54}(\ln(n+1)+\gamma)+\dfrac{1}{54}(1+\tfrac12+\tfrac13)\\ \approx -\dfrac{11}{864}(\ln(n-2)+\gamma)\\ \approx -\dfrac{11}{864}(\ln(n+2)+\gamma)+\dfrac{11}{864}(1+\tfrac12+\tfrac13+\tfrac14)\\ +\dfrac{1}{36}\dfrac{\pi^2}{6}-\dfrac{1}{36}\\ -\dfrac{1}{36}\dfrac{\pi^2}{6}+\dfrac{1}{36}(1+\tfrac14+\tfrac19)\\ -\dfrac{1}{288}\dfrac{\pi^2}{6}+\dfrac{1}{288}(1+\tfrac14+\tfrac19+\tfrac{1}{16})\\ +\dfrac{1}{288}\dfrac{\pi^2}{6} \end{cases}$$

Happily, all "non rational terms" (containing $\log N$, $\gamma, \frac{\pi^2}{6}$) asymptoticaly cancel ; it remains only rationals summing up to the same result $1/4608$ as found by @Macavity (a big thank to him for having spotted an error of mine).

Important remark: in fact, partial sum $S_n$ (defined in (1)) can be given a closed form expression (using Wolfram Alpha again):

$$S_n=\dfrac{(n-2)(n+3) (n^2+n+4)(n^4+2 n^3 - n^2 -2n + 24)}{4608(n-1)^2 n^2 (n+1)^2 (n+2)^2}$$

from which the limit is evident!

Jean Marie
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    @macavity A big thank for having spotted the errors. – Jean Marie Apr 16 '21 at 11:17
  • @macavity See the remark I just added at the end of my answer. – Jean Marie Apr 16 '21 at 12:31
  • Saw that, thanks. of course the partial sums are well defined rational ones, and easily written down by telescoping as well.. – Macavity Apr 16 '21 at 13:02
  • @Macavity I was surprized by the fact that the denominator of the partial sum was "mimicking" the denominator of the general term. Do you know where one can find a rather general proof of this fact ? Maybe in "Concrete Mathematics" ? – Jean Marie Apr 16 '21 at 14:27
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    That may not be always true, though certainly when the sum telescopes it has to hold true. If the general term can be written as $f(n)-f(n+1)$ where $f\to 0$, partial sum will obviously be $f(3)-f(n)$ if summed from $3$ to $\infty$ is taken. As $f(3)$ is just a number, bringing to a common denominator preserves the denominator of $f(n)$. – Macavity Apr 16 '21 at 14:51
  • @macavity This makes sense. Somewhat a criteria for a sum to be telescopic...Thank you very much, once again. – Jean Marie Apr 16 '21 at 15:33