What is a good estimate for $\sum_{t=0}^T \frac{1}{\gamma^t}\frac{1}{\sqrt{t + 1}}$, where $0 < \gamma < 1$?

Question

Let $0 < \gamma < 1$ and $T$ be a "large" nonnegative integer.

Question

What is a good estimate (upper bound) for $\sum_{t=0}^T\dfrac{1}{\gamma^t}\dfrac{1}{\sqrt{t + 1}}$?

In general, for $\alpha > 1$, I'm interested in estimating

$\sum_{t=0}^T\dfrac{1}{\gamma^t}\dfrac{1}{(t + 1)^{1/\alpha}}$

Answer 1

According to Wolfram Alpha, we can express (with $T = \infty$), \begin{align*} \sum_{t=0}^{\infty} \frac{1}{\gamma^t(t+1)^{1/\alpha}} = \gamma \text{Li}_{1/\alpha}\left(\frac{1}{\gamma}\right) \end{align*} where $\text{Li}_n(x)$ is the polylogarithm function. If you are looking for strict bounds in terms of elementary functions, then something like Holder's inequality gives us \begin{align*} \sum_{t=0}^{T} \frac{1}{\gamma^t(t+1)^{1/\alpha}} &\le \left(\sum_{t=0}^{T}\frac{1}{\gamma^{t\alpha/(\alpha-1)}}\right)^{1 - 1/\alpha}\left(\sum_{t=0}^{T}\frac{1}{t+1}\right)^{1/\alpha} \end{align*} The first product on the RHS is a partial geometric series (with a closed-form). and the second is $H_{T+1}^{1/\alpha}$, where $H_n$ is the $n$th harmonic number, and this can be approximated very well with $\log$ and the Euler-Mascheroni constant.

EDIT: In Holder's, I applied with $p = \alpha/(\alpha-1)$ and $q = \alpha$. The resulting partial harmonic series doesn't converge, so the bound is quite weak. Applying with $p = 2\alpha/(2\alpha-1)$ and $q = 2\alpha$ will solve this issue.