$$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}$$$

Find $$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}$$$.

Perform partial fraction decomposition (for steps, see partial fraction decomposition calculator)::

$${\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}\right)}}={\color{red}{\left(\sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} + \frac{1}{16 \left(2 k + 1\right)^{2}} - \frac{1}{32 \left(2 k - 3\right)} + \frac{1}{16 \left(2 k - 3\right)^{2}}\right)\right)}}$$

Split the series:

$${\color{red}{\left(\sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} + \frac{1}{16 \left(2 k + 1\right)^{2}} - \frac{1}{32 \left(2 k - 3\right)} + \frac{1}{16 \left(2 k - 3\right)^{2}}\right)\right)}}={\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{16 \left(2 k + 1\right)^{2}} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)\right)}}$$

Shift the series by $$$1$$$:

$${\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{16 \left(2 k + 1\right)^{2}}\right)}} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)={\color{red}{\left(\sum_{k=2}^{\infty} \frac{1}{16 \left(2 k - 1\right)^{2}}\right)}} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Pull the constant out of the series:

$${\color{red}{\left(\sum_{k=2}^{\infty} \frac{1}{16 \left(2 k - 1\right)^{2}}\right)}} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)={\color{red}{\left(\frac{\sum_{k=2}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}}{16}\right)}} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Split the series:

$$\frac{{\color{red}{\left(\sum_{k=2}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=\frac{{\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}} + \sum_{k=1}^{1} - \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms.

$$\frac{{\color{red}{\left(\sum_{k=1}^{1} - \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \frac{\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}}{16} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=\frac{{\color{red}{\left(-1\right)}}}{16} + \frac{\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}}{16} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

$$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}$$$ is a known series.

It is $$$\sum_{k=1}^{\infty} \left(2 k - 1\right)^{- n_{0}}=\left(1 - 2^{- n_{0}}\right) \zeta\left(n_{0}\right)$$$, $$$n_{0} > 1$$$ with $$$n_{0}=2$$$.

Therefore,

$$- \frac{1}{16} + \frac{{\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=- \frac{1}{16} + \frac{{\color{red}{\left(\frac{\pi^{2}}{8}\right)}}}{16} + \sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Shift the series by $$$-1$$$:

$$- \frac{1}{16} + \frac{\pi^{2}}{128} + {\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{16 \left(2 k - 3\right)^{2}}\right)}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=- \frac{1}{16} + \frac{\pi^{2}}{128} + {\color{red}{\left(\sum_{k=0}^{\infty} \frac{1}{16 \left(2 k - 1\right)^{2}}\right)}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Pull the constant out of the series:

$$- \frac{1}{16} + \frac{\pi^{2}}{128} + {\color{red}{\left(\sum_{k=0}^{\infty} \frac{1}{16 \left(2 k - 1\right)^{2}}\right)}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=- \frac{1}{16} + \frac{\pi^{2}}{128} + {\color{red}{\left(\frac{\sum_{k=0}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}}{16}\right)}} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Split the series:

$$- \frac{1}{16} + \frac{{\color{red}{\left(\sum_{k=0}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \frac{\pi^{2}}{128} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=- \frac{1}{16} + \frac{{\color{red}{\left(\sum_{k=0}^{0} \frac{1}{\left(2 k - 1\right)^{2}} + \sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \frac{\pi^{2}}{128} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms.

$$- \frac{1}{16} + \frac{{\color{red}{\left(\sum_{k=0}^{0} \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \frac{\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}}{16} + \frac{\pi^{2}}{128} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=- \frac{1}{16} + \frac{{\color{red}{1}}}{16} + \frac{\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}}{16} + \frac{\pi^{2}}{128} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

$$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}$$$ is a known series.

It is $$$\sum_{k=1}^{\infty} \left(2 k - 1\right)^{- n_{0}}=\left(1 - 2^{- n_{0}}\right) \zeta\left(n_{0}\right)$$$, $$$n_{0} > 1$$$ with $$$n_{0}=2$$$.

Therefore,

$$\frac{{\color{red}{\left(\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 1\right)^{2}}\right)}}}{16} + \frac{\pi^{2}}{128} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)=\frac{{\color{red}{\left(\frac{\pi^{2}}{8}\right)}}}{16} + \frac{\pi^{2}}{128} + \sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)$$

Steps are not available for this series.

$$\frac{\pi^{2}}{64} + {\color{red}{\left(\sum_{k=1}^{\infty} \left(\frac{1}{32 \left(2 k + 1\right)} - \frac{1}{32 \left(2 k - 3\right)}\right)\right)}}=\frac{\pi^{2}}{64} + {\color{red}{\left(0\right)}}$$

Hence,

$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}}=\frac{\pi^{2}}{64}$$

$$$\sum_{k=1}^{\infty} \frac{1}{\left(2 k - 3\right)^{2} \left(2 k + 1\right)^{2}} = \frac{\pi^{2}}{64}\approx 0.154212568767021$$$A