One estimate about series

\mathbf{Problem:} Suppose \rho>0, u\in (0,1), then \sum\limits_{n=1}^\infty n^\rho u^n=O((1-u)^{-\rho-1}) as u\to 1+.

\mathbf{Proof:} \forall \,n\geq 1, n\leq x<n+1, we have

\displaystyle \frac{x^\rho u^x}{n^\rho u^n}=(\frac{x}{n})^\rho u(x-n)\geq u

So
\displaystyle \sum\limits_{n=1}^\infty n^\rho u^n\leq \frac 1u \sum\limits_{n=1}^\infty \int_n^{n+1} x^\rho u^x=\frac 1u \int_1^\infty x^\rho u^x dx=\frac{1}{u(-\log u)^{\rho+1}}\int_{-\log u}^\infty t^\rho e^{-t}dt

This means \displaystyle\sum\limits_{n=1}^\infty n^\rho u^n\leq \frac{1}{u(-\log u)^{\rho+1}} \Gamma (\rho+1).
Since we have \frac{1}{-\log u}=O(\frac{1}{1-u}) as u\to 1+, then \sum\limits_{n=1}^\infty n^\rho u^n=O((1-u)^{-\rho-1}) as u\to 1+.

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