9
$\begingroup$

I have been trying to figure out how to evaluate the following sum: $$S_n=\sum_{k=1}^{n} \frac{k}{k^4+1/4}$$

In the problem, the value of $S_{10}$ was given as $\frac{220}{221}$.

I have tried partial decomposition, no where I go. Series only seems like it telescopes, otherwise there isn't another way.

Any ideas are appreciated!

$\endgroup$
  • $\begingroup$ the solution containes the PolyGamma function $\endgroup$ – Dr. Sonnhard Graubner Sep 28 '17 at 15:03
  • $\begingroup$ @Dr.SonnhardGraubner Thanks, can you please explain me about it more? $\endgroup$ – akhmeteni Sep 28 '17 at 15:03
  • $\begingroup$ In general, any quartic polynomial of the form $x^4 + bx^2 + cx + d$ (there is a missing $x^3$ term) can be factored into $(x^2 + mx + p)(x^2 - mx + q)$ $\endgroup$ – DanielV Sep 29 '17 at 23:33
  • $\begingroup$ @JackD'Aurizio Where is the $q$ in the right hand side of the equation? If my old old textbook is correct then this formula (or at least this approach to factoring quartics) is due to Pascal. $\endgroup$ – DanielV Oct 1 '17 at 19:28
  • $\begingroup$ @DanielV: oh, sorry, I misread it. $\endgroup$ – Jack D'Aurizio Oct 1 '17 at 19:29
18
$\begingroup$

By Sophie Germain's identity $$ 4k^4+1 = (2k^2+2k+1)(2k^2-2k+1) \tag{A}$$ hence $$ \frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} = \frac{4k}{4k^4+1} = \frac{k}{k^4+1/4}\tag{B} $$ and we may notice that by setting $p(x)=2x^2-2x+1$ we have $p(x+1)=2x^2+2x+1$.
In particular $$ \sum_{k=1}^{n}\frac{k}{k^4+1/4}=\sum_{k=1}^{n}\left(\frac{1}{p(k)}-\frac{1}{p(k+1)}\right) = \frac{1}{p(1)}-\frac{1}{p(n+1)}=1-\frac{1}{2n^2+2n+1} $$ equals $\frac{2n^2+2n}{2n^2+2n+1}$ for any $n\geq 1$.


Telescoping is not strictly necessary to be able to compute the value of similar series. For instance $$ \sum_{k\geq 0}\frac{1}{k^4+4} = \frac{\pi\cos\pi+\sinh\pi}{8\sinh\pi}, $$ but this is a different story, related with Weierstrass products, the Poisson summation formula or the (inverse) Laplace transform.

$\endgroup$
  • $\begingroup$ Thanks for enlightening answer! I may ask how does one get hints on thinking in that direction? $\endgroup$ – akhmeteni Sep 28 '17 at 15:15
  • $\begingroup$ Also tell me where I can find more identities like the one you provided (Sophie-germain identity)! $\endgroup$ – akhmeteni Sep 28 '17 at 15:17
  • 1
    $\begingroup$ @akhmeteni: I can only give you the hard answer: experience. Practice makes perfect, meaning that it really helps you in recognizing which paths are plausible and which paths are not. $\endgroup$ – Jack D'Aurizio Sep 28 '17 at 15:18
  • $\begingroup$ @akhmeteni: about SGI: artofproblemsolving.com/wiki/… $\endgroup$ – Jack D'Aurizio Sep 28 '17 at 15:18
  • 2
    $\begingroup$ +1 for the very simple telescoping approach and the remark at the end is a big bonus. $\endgroup$ – Paramanand Singh Oct 1 '17 at 14:21
4
$\begingroup$

Try to break the denominator into product of two factors:

$$\begin{align} 4k^4 + 1 &= (2k^2)^2 + 1 + 2 (2k^2) - 2 (2k^2) \\ &= (2k^2 +1)^2 - (2k)^2 \\ &= (2k^2 +2k +1)(2k^2 -2k+1) \end{align}$$

Using this we see the general term as:

$$T_k = \dfrac{1}{2k^2-2k+1} - \dfrac{1}{2k^2+2k+1} \\ T_{k+1} = \dfrac{1}{2k^2+2k+1} - \dfrac{1}{2(k+1)^2+2(k+1)+1} $$

Alternate terms cancel and the sum telescopes to:

$$1-\frac{1}{2n^2+2n+1}$$

$\endgroup$
  • $\begingroup$ Thanks to you too! $\endgroup$ – akhmeteni Sep 28 '17 at 15:23
  • $\begingroup$ you are welcome :) $\endgroup$ – samjoe Sep 28 '17 at 15:24
1
$\begingroup$

The solutions by Jack d'Aurizio and samjoe are no doubt optimal, but I wondered if one could discover a solution even if one were not so clever as to see the (Sophie Germain) factorization. Here are the values of $S_n = \sum_{k=1}^{n}\frac{k}{k^4+1/4}$ for $n \le 15$: $$\frac{4}{5},\frac{12}{13},\frac{24}{25},\frac{40}{41},\frac{60}{61},\frac{84}{85}, \frac{112}{113},\frac{144}{145},\frac{180}{181},\frac{220}{221},\frac{264}{265},\frac{ 312}{313},\frac{364}{365},\frac{420}{421},\frac{480}{481}. $$ One notices that the numerators are divisible by 4; dividing by 4, one recognizes the sequence of triangular numbers, $t_n = n(n+1)/2$. Thus one can guess that $S_n = \frac{4 t_n}{4 t_n + 1}$. Then one can set out to prove this guess by induction. The induction step requires proving the identity $$\frac{4 t_n}{4 t_n + 1} + \frac{4(n+1)}{4(n+1)^4 + 1} = \frac{4 t_{n+1}}{4 t_{n+1} + 1},$$ or equivalently $$\frac{4(n+1)}{4(n+1)^4 + 1} = \frac{4 t_{n+1}}{4 t_{n+1} + 1} -\frac{4 t_n}{4 t_n + 1}.$$ It is straightforward to verify this by simplifying the RHS, and in the course of this, one discovers the Sophie Germain factorization.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.