Prove that the series $\sum_{n=1}^\infty \frac{24n^3 - 4n^2 - 9n + 3}{n^2 (2n-1)^2 (n+1)}$ converges to 11

Question

I am studying the following infinite series:

$$\sum_{n=1}^\infty \frac{24n^3 - 4n^2 - 9n + 3}{n^2 (2n-1)^2 (n+1)}$$

Numerical approximations suggest that this series converges to a value very close to 11. However, I am unable to provide a formal proof of this result. Using numerical computations, I calculated partial sums for increasingly large $n$:

• for $n = 100$, $\approx 10.76$,
• for $n = 200$, $\approx 10.79$,
• for $n = 500$, $\approx 10.81$,

and so on. Using high-precision arithmetic, the partial sums approach 11 but do not seem to reach it exactly, likely due to numerical limitations.

Could someone provide guidance or an analytical approach to prove that this series converges to 11? Any suggestions — whether through known summation techniques, connections to other series, or asymptotic analysis — would be greatly appreciated.

Thank you for your time and insights!

Answer 1

$$S=\sum_{n=1}^\infty \frac{24n^3 - 4n^2 - 9n + 3}{n^2 (2n-1)^2 (n+1)}$$

Perform partial fractions,

$$\color{blue}{\frac{24 n^{3} - 4 n^{2} - 9 n + 3}{n^{2} \left(n + 1\right) \left(2 n - 1\right)^{2}}=\frac{3}{n^{2}}+\frac{\frac{32}{9}}{2 n - 1}+\frac{\frac{4}{3}}{\left(2 n - 1\right)^{2}}+\frac{- \frac{16}{9}}{n + 1}}$$

$$\frac{24 n^{3} - 4 n^{2} - 9 n + 3}{n^{2} \left(n + 1\right) \left(2 n - 1\right)^{2}}=3\frac{1}{n^{2}}+\frac{32}{3}\frac{1}{(2 n - 1)(2n+2)}+\frac{4}{3}\frac{1}{\left(2 n - 1\right)^{2}}$$

$$S=3\sum_{n=1}^\infty\frac{1}{n^{2}}+\frac{32}{3}\sum_{n=1}^\infty\frac{1}{(2 n - 1)(2n+2)}+\frac{4}{3}\sum_{n=1}^\infty\frac{1}{\left(2 n - 1\right)^{2}}$$

• $\sum_{n=1}^\infty\frac{1}{n^{2}} = \zeta(2)$

• $\sum_{n=1}^\infty\frac{1}{(2 n - 1)(2n+2)}=\frac{\log(2)}{3}+\frac16$

• $\sum_{n=1}^\infty\frac{1}{\left(2 n - 1\right)^{2}}=\frac{\pi^2}{8}$

$$\boxed{\sum_{n=1}^\infty \frac{24n^3 - 4n^2 - 9n + 3}{n^2 (2n-1)^2 (n+1)}=\frac{2}{3}\pi^2+\frac{16}{9}+\frac{32}{9}\log(2)}$$

$$S\approx 10.822<11$$

Useful References

• Reference 1

• Reference 2

• Reference 3

• Reference 4

• Reference 5

• Reference 6

• Reference 7

• Reference 8

• Reference 9

Answer 2

Since it looks like you are new to these stuff, I would like to make a quick review for you about Infinite Series.

Check out Partial fraction along with Gauss's digamma theorem and special values of PolyGamma. Or if you sum over all integers, you can use Residue theorem or this specific technique (proved by inverse Laplace transform):
$$f_{a,b}(n)=\frac{e^{ibn}}{(an)^2+1}=\frac{1}{2}\int_{-\infty}^{\infty}e^{-|x|}e^{in(ax+b)}dx$$
This has the advantage of being multiplicative in case you have multiple cells of $f_{a,b}(n)$. Combine this with Dirac comb: $\sum_{k=-N}^{N}e^{iky}\to2\pi\sum_{k\in\mathbb{Z}}\delta(y-2\pi k)$ (where $\delta(x)$ is Delta Dirac distribution) can produce geometric sums. One example is here

One special trick is that you can "modify" the sum so that you can sum over all integers, for example:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\lim_{h\to0}\frac{1}{2}\left(-\frac{1}{h^2}+\sum_{n=-\infty}^{\infty}\frac{1}{n^2+h^2}\right)$$
Also, there are a lot of beautiful sums on MSE here, this is one of my favorites

Personal advice: Don't trust numerical results that way. Please consider using Mathematica. Your sum's value is $\frac{2}{9}(8+3\pi^2+8\log(4))=10.822...$, not $11$.

Answer 3

Let $A:=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$ and $B:=\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2.$ $$\frac{24n^3-4n^2-9n+3}{n^2(2n-1)^2(n+1)}=\frac3{n^2}+\frac{4/3}{(2n-1)^2}+\frac{32}9\left(\frac1{2n-1}-\frac1{2n+2}\right)$$ hence $$\begin{align}\sum_{n=1}^\infty\frac{24n^3-4n^2-9n+3}{n^2(2n-1)^2(n+1)}&=3A+\frac43\left(A-\frac A4\right)+\frac{32}9\left(B+\frac12\right)\\&=4A+\frac{32}9B+\frac{16}9\\ &=\frac{2\pi^2}3+\frac{32}9\ln2+\frac{16}9\\&\approx10.822. \end{align}$$

Answer 4

You could even have a good approximation of the partial sum

$$S_p=\sum_{n=1}^p \frac{24n^3 - 4n^2 - 9n + 3}{n^2 (2n-1)^2 (n+1)}$$ After partial fraction decomposition, you have a sum of polygamma functions and using their asymptotics $$S_p=\frac{2}{9} \left(8+3 \pi ^2+16 \log (2)\right)-\frac{6}{p}+\frac{7}{2 p^2}-\frac{9}{4 p^3}+O\left(\frac{1}{p^4}\right)$$ which is in a relative error smaller than $0.10$% if $p=4$ and smaller than $0.01$% if $p=7$.