5

I was computing a series and solved until $$\sum_{k=1}^\infty \frac {1}{((2k-1)(2k+1))^2}$$

I know Telescopic series but this doesn’t seem like it can be transform into that.

Upon solving this I got $$\frac{1}{4} \sum_{k=1}^\infty \frac {1}{(2k-1)^2} + \frac {1}{(2k+1)^2} + \frac {1}{(2k+1)} - \frac {1}{(2k-1)} $$

Samar
  • 556
  • How have you tried approaching the problem? Mentioning "telescopic series" suggests you've considered a partial fraction expansion, maybe - you should include whatever progress you've made in the body of your question. – user170231 Apr 07 '23 at 16:26
  • This is not a simple telescopic series.... it's way more interesting – Blue Cat Blues Apr 07 '23 at 16:26
  • 1
    Solve partial fractions: $$\frac1{((2k-1)(2k+1))^2}=\frac{a}{(2k+1)^2}+\frac b{(2k-1)^2}+\frac c{2k+1}+\frac d{2k-1}$$ and you might simplify the sum into pieces you can sum. – Thomas Andrews Apr 07 '23 at 16:26
  • Recall the power series$$\tanh^{-1}(x)=\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}$$ – user170231 Apr 07 '23 at 16:52
  • @user170231 That series is irrelevant: the linear terms here do telescope, and the series you've given diverges at $1$. – JBL Apr 07 '23 at 16:56
  • @JBL Not irrelevant, it can be used to address the sums with squared denominators. – user170231 Apr 07 '23 at 16:57

3 Answers3

3

Let's try to solve this using partial fraction,

we can rewrite it as, \begin{align*} \sum_{k=1}^\infty \frac{1}{(2k-1)^2 \, (2k+1)^2} &= \frac{1}{2} \sum_{k=1}^\infty \left( \frac{1}{(2k-1)^2 \, (2k+1)} - \frac{1}{(2k-1) \, (2k+1)^2} \right) \\ &= \frac{1}{4} \sum_{k=1}^\infty \left( \frac{1}{(2k-1)^2} - \frac{1}{(2k-1) \, (2k+1)} - \frac{1}{(2k-1) \, (2k+1)} + \frac{1}{(2k+1)^2} \right) \\ &= \frac{1}{4} \sum_{k=1}^\infty \frac{1}{(2k-1)^2} + \frac{1}{4} \sum_{k=1}^\infty \frac{1}{(2k+1)^2} - \frac{1}{2} \sum_{k=1}^\infty \frac{1}{(2k-1)\, (2k+1)} \end{align*}

Now first two summations can be calculated in terms of $\zeta$ function and last one is just telescoping series.

See we know that \begin{align*} \zeta(2) &= \sum_{k=1}^\infty \frac{1}{k^2} \\ &= \left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots \right) + \left( \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots \right) \\ &= \sum_{k=1}^\infty \frac{1}{(2k-1)^2} + \frac{1}{2^2} \zeta(2) \end{align*}

So $$ \sum_{k=1}^\infty \frac{1}{(2k-1)^2} = \frac{3}{4} \zeta(2) = \frac{\pi^2}{8} $$

Likewise you can calculate the 2nd one also :)

  • Thanks for your efforts but I first have to learn about zeta function. I was just doing this series for fun, I didn’t know that this requires zeta function. – Samar Apr 07 '23 at 17:09
  • @SamarKamboj yeah! Zeta function requires here to calculate $1/1^2 + 1/2^2 + \cdots$ Anyway, learn about this cool function. Have a mathematical day! –  Apr 07 '23 at 18:32
3

$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2} \left(2 k + 1\right)^{2}}=\sum_{k=1}^{\infty} \frac{1}{16 k^{4} - 8 k^{2} + 1}$$ $$={\sum_{k=1}^{\infty} \left(\frac{1}{4 \left(2 k + 1\right)} + \frac{1}{4 \left(2 k + 1\right)^{2}} - \frac{1}{4 \left(2 k - 1\right)} + \frac{1}{4 \left(2 k - 1\right)^{2}}\right)}$$ Now $$\frac14\sum_{k=1}^{\infty}\left(\frac{1}{2k+1}-\frac{1}{2k-1}\right)$$ is a simple telescopic series whose value is equal to $$\color{blue}{\frac{-1}{4}}$$ We're left with $$\frac14\sum_{k=1}^{\infty}\left(\frac{1}{(2k+1)^2}+\frac{1}{(2k-1)^2}\right)$$ Recall that $$\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}$$ is a known series. Its general form is $(n_0>1)$ $$\sum_{k=1}^{\infty} \left(2 k - 1\right)^{- n_{0}}=\left(1 - 2^{- n_{0}}\right) \zeta\left(n_{0}\right)$$ where we have $n_0=2$

So $$\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$$

Now we're left with $$\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$$ Now if $$\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$$ then automatically $$\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}=\frac{\pi^2}{8}-1$$ as it contains one less term than the previous series.

So $$\frac14\sum_{k=1}^{\infty}\left(\frac{1}{(2k+1)^2}+\frac{1}{(2k-1)^2}\right)=\frac14\left(\frac{\pi^2}{8}+\frac{\pi^2}{8}-1\right)$$ $$=\frac{\pi^2}{16}-\frac14$$ $$=\frac{\pi^2-4}{16}$$ Adding the $\frac{-1}{4}$ of the previous telescopic sum to it gives the final answer $$\boxed{\color{red}{\frac{\pi^2-8}{16}}}$$

Blue Cat Blues
  • 10,058
  • Thank you for your time to write this answer. Honestly I got this from AoPS High School Math community. I didn’t know that this requires zeta function. I don’t know about zeta function. I just know one series similar to this one which is $\sum_{k=1}^\infty \frac {1}{k^2}$ – Samar Apr 07 '23 at 17:05
  • In my honest opinion, this is way above high school level...but it's good that atleast you're trying to solve these kind of problems... – Blue Cat Blues Apr 07 '23 at 17:30
3

Expanding into partial fractions yields, as you've shown,

$$S = \sum_{k=1}^\infty \frac1{(2k-1)^2(2k+1)^2} = \frac14 \sum_{k=1}^\infty \left[\underbrace{\frac1{2k+1} - \frac1{2k-1}}_{\rm telescoping} + \underbrace{\frac1{(2k+1)^2} + \frac1{(2k-1)^2}}_{S'}\right]$$

The telescoping part converges to $-1$. For the non-telescoping part, recall the power series

$$\tanh^{-1}(x) = \sum_{k=0}^\infty \frac{x^{2k+1}}{2k+1}$$

Now let

$$f(x) = \sum_{k=1}^\infty \frac{x^{2k+1}}{(2k+1)^2} \quad \text{and} \quad g(x) = \sum_{k=1}^\infty \frac{x^{2k-1}}{(2k-1)^2} = \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)^2} = x + f(x)$$

Differentiate with respect to $x$, multiply by $x$, apply the fundamental theorem of calculus to find $f$ and $g$, and let $x\to1$ from below to recover $S'$.

$$\begin{align*} f'(x) = \sum_{k=1}^\infty \frac{x^{2k}}{2k+1} \implies x f'(x) &= \tanh^{-1}(x) - x \\ \implies f(x) &= f(0) + \int_0^x \frac{\tanh^{-1}(y)-y}y \, dy \\ \implies \sum_{k=1}^\infty \frac1{(2k+1)^2} &= \lim_{x\to1^-} f(x) = \int_0^1 \frac{\tanh^{-1}(y)-y}y \, dy \end{align*}$$

Now,

$$S = \frac{S'-1}4 = - \frac12 + \frac12 \int_0^1 \frac{\tanh^{-1}(y)}y \, dy$$

Integrate by parts to write

$$\int \frac{\tanh^{-1}(y)}y \, dy = \tanh^{-1}(y) \ln(y) - \int \frac{\ln(y)}{1-y^2} \, dy$$

and consult here for methods to compute the remaining integral.

user170231
  • 25,320