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Can someone give a hint on how to show that for all $k > 1$, $\exists \lambda$, so that \begin{align*} \sum_{m = c + 1}^{\infty} \frac{m^{-k}}{(m-c)^2} \leq \lambda c^{-k} \end{align*} And find an expression of $\lambda(k)$ that realizes this. This expression is, of course, the smaller the better

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Rewriting your sum can be written as $$\sum_{m=1}^{\infty} \dfrac1{(c+m)^k m^2} = c^{-k} \sum_{m=1}^{\infty} \dfrac1{m^2(1+m/c)^k} < c^{-k} \sum_{m=1}^{\infty} \dfrac1{m^2} = \zeta(2)c^{-k}$$

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The given sum can be upper bounded by $$\sum_{m\ge c+1}m^{-k}\sum_{m\ge c+1}\frac{1}{(m-c)^2}=\lambda(k)c^{-k}$$ where $\lambda(k)=\zeta(2)c^k\sum_{m\ge c+1}m^{-k}=\zeta(2)c^k\left(\zeta(k)-\sum_{j=1}^c j^{-k}\right)$

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