# How to evaluate the sum : $\sum_{k=1}^{n} \frac{k}{k^4+1/4}$

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!

• the solution containes the PolyGamma function – Dr. Sonnhard Graubner Sep 28 '17 at 15:03
• @Dr.SonnhardGraubner Thanks, can you please explain me about it more? – akhmeteni Sep 28 '17 at 15:03
• 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 $(x^2 + mx + p)(x^2 - mx + q)$ – DanielV Sep 29 '17 at 23:33
• @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. – DanielV Oct 1 '17 at 19:28
• @DanielV: oh, sorry, I misread it. – Jack D'Aurizio Oct 1 '17 at 19:29

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.

• Thanks for enlightening answer! I may ask how does one get hints on thinking in that direction? – akhmeteni Sep 28 '17 at 15:15
• Also tell me where I can find more identities like the one you provided (Sophie-germain identity)! – akhmeteni Sep 28 '17 at 15:17
• @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. – Jack D'Aurizio Sep 28 '17 at 15:18
• @akhmeteni: about SGI: artofproblemsolving.com/wiki/… – Jack D'Aurizio Sep 28 '17 at 15:18
• +1 for the very simple telescoping approach and the remark at the end is a big bonus. – Paramanand Singh Oct 1 '17 at 14:21

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:

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

• Thanks to you too! – akhmeteni Sep 28 '17 at 15:23
• you are welcome :) – SJ. Sep 28 '17 at 15:24

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.