## Only A Test

This is simply a test of the Latex to WordPress converter available here. $\displaystyle -\sum_{n=1}^\infty \frac{\log n}{n} \times \left\{-1,-1,\cdots,k-1,-1,-1,\cdots,k-1,\cdots \right\}$

where the term ${k-1}$ appears at ${n=0 \bmod k}$.

We can rewrite this as $\displaystyle -\frac{\partial}{\partial \nu} \sum_{m=1}^\infty \left[ \sum_{n=1}^{k-1} \frac{1}{(n +k(m-1))^\nu} + (1-k)\frac{1}{(m k)^\nu} \right] \vert_{\nu=1}$ $\displaystyle = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k-1} \sum_{m=1}^\infty \frac{1}{ k^\nu} \frac{1}{(m-1+\frac{n}{k})^\nu} \right] \vert_{\nu=1}$ $\displaystyle = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k-1} \frac{1}{ k^\nu} \sum_{m=0}^\infty \frac{1}{(m+\frac{n}{k})^\nu} \right] \vert_{\nu=1}$ $\displaystyle = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k-1} \frac{1}{ k^\nu} \hat{\zeta} \left(\nu,\frac{n}{k}\right) \right] \vert_{\nu=1}$ $\displaystyle = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \sum_{n=1}^{k} \frac{1}{ k^\nu} \hat{\zeta} \left(\nu,\frac{n}{k}\right) - \hat{\zeta}(\nu,1) \right] \vert_{\nu=1}$ $\displaystyle = -\frac{\partial}{\partial \nu} \left[ \frac{1-k}{k^\nu}\zeta(\nu) + \zeta(\nu) - \hat{\zeta}(\nu,1) \right] \vert_{\nu=1}$ $\displaystyle = -k^{-\nu} \left[ k \log k \zeta(\nu) + (k^\nu - k) \zeta'(\nu) \right] \vert_{\nu=1}$

Now use the series, for ${z}$ near 1, $\displaystyle \zeta(z) \approx \frac{1}{z-1} + \gamma -\gamma_1 (z-1)$

and put ${\nu = 1 + \epsilon}$ to obtain $\displaystyle = -\frac{1}{k} \left[ k \log k \zeta(1+\epsilon) + k(\epsilon \log k + \frac{1}{2} \epsilon^2 \log^2 k ) \zeta'(1+\epsilon) \right]_{\epsilon=0}$ $\displaystyle = -\gamma \log k + \frac{1}{2} \log^2 k$

as desired.