Tail of a divergent $p$-series with $p<1$.

Question

Consider a divergent $p$-series with $p<1$, for instance $\sum_{j=0}^{\infty} \frac{1}{\sqrt{j}}$. Let $R_{n}=\sum_{j=n}^{\infty} \frac{1}{\sqrt{j}}$ be the tail of this series. Obviously $R_n$ diverges as $n \to \infty$, so I was ridiculously trying to find the rate of divergence of the tail. I thought that since the tail of harmonic series behaves very well, we can find the rate $R_n$ as $O(n^{\alpha})$ for some $\alpha>0$. At first I thought that $\alpha$ could be $1/2$, which is clearly incorrect. Is there any way to find the rate of divergence of this sum? Seems my knowledge in approximation is poor. :(

Answer 1

The "tail" $$ R_n(\alpha) = \sum_{k=n}^\infty \frac{1}{k^{\alpha}} $$ for $\alpha \leq 1$ will not be very interesting to study: it does not "diverge" to infinity, it is infinite (exactly because... the series diverges to infinity).

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However, a more interesting question would be the rate of divergence of the partial sums $$ S_n(\alpha) = \sum_{k=1}^n \frac{1}{k^{\alpha}} $$ for $\alpha \leq 1$. Here, a comparison series-integral shows that $$ S_n(\alpha) \operatorname*{\sim}_{n\to\infty} \int_1^n \frac{dx}{x^\alpha} $$ so that $$ S_n(\alpha) \operatorname*{\sim}_{n\to\infty} \begin{cases}\ln n & \alpha=1\\\frac{x^{1-\alpha}}{(1-\alpha)} & \alpha < 1\end{cases} $$ For instance, for $\alpha=1/2$, you get $$\sum_{k=1}^n \frac{1}{\sqrt{n}} \operatorname*{\sim}_{n\to\infty} 2\sqrt{n}\,.$$