$$\sum_{n=1}^{ \infty}{\frac{1}{n^2(8n+1)^2}}$$
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MathBot Answer:
\[\sum_{n=1}^{\infty} \frac{1}{n^{2} \left(8 n + 1\right)^{2}} = - 16 \gamma + \psi^{(1)}\left(\frac{9}{8} \right) + \frac{\pi^{2}}{6} - 16 \psi \left(\frac{9}{8} \right) \approx 0.013499\]
$$\begin{aligned}\sum_{n=1}^{\infty} \frac{1}{n^{2} \left(8 n + 1\right)^{2}} &= \sum_{n=1}^{\infty} \left(\frac{128}{\left(8 n + 1\right)} + \frac{64}{\left(8 n + 1\right)^{2}} - \frac{16}{n} + \frac{1}{n^{2}}\right)\\&= - \left(-16\right) \psi \left(1 + 0 \right) + 1 \psi^{(1)}\left(1 + 0 \right) + 1 \psi^{(1)}\left(1 + \frac{1}{8} \right) - 16 \psi \left(1 + \frac{1}{8} \right) \ \ \ \small{\color{grey}\text{using the formula } \sum_{n=1}^\infty u(n) = (-1)^{d} \sum_{j=1}^m\frac{a_j}{(d-1)!} \psi^{(d-1)}\left(1+\alpha_j\right)\text{where } u(n) = \frac{p(n)}{q(n)} = \sum_{k=1}^{m} \frac{a_k}{(n+\alpha_k)^d}}\\&= - 16 \gamma + \frac{\pi^{2}}{6} - 16 \psi \left(\frac{9}{8} \right) + \psi^{(1)}\left(\frac{9}{8} \right) \ \ \ \small{\color{grey}\text{using the formula } \psi^{(m)}(z) = \frac{d^{(m+1)}}{dz^{(m+1)}}\ln\Gamma(z)}\\&\approx 0.013499\end{aligned}$$
\(\gamma\) is the Euler-Mascheroni constant, sometimes called Euler's constant, a mathematical constant that is approximately \(0.577216\).
\(\psi^{(m)}\) is the polygamma function of order \(m\), a special function given by the \(mth\) derivative of the digamma function. It is defined by the formula \(\psi^{(m)}(z) = \frac{d^{(m+1)}}{dz^{(m+1)}}\ln\Gamma(z)\).
\(\psi\) is the digamma function, a special function given by the logarithmic derivative of the gamma function. It is defined by the formula \(\psi(z) = \frac{d}{dz}\ln\Gamma(z) = \frac{\Gamma'(z)}{\Gamma(z)}\).