Show that $ \sum_{j=1}^{k} \frac{\log {p_j}}{p_{k+1}^s}$ converges to 0

Question

Notation: Let $p_0=1$, $p_1=2$, $p_2=3, \cdots, p_k$ be succesive primes.

Is $$ \lim_{k\rightarrow \infty} \,\,\sum_{j=1}^{k} \frac{\log {p_j}}{p_{k+1}^s} = \lim_{k\rightarrow \infty} \,\,\frac{1}{p_{k+1}^s}\log \left[ \prod_{j=1}^{k}{p_j}\right] =0$$ where $s \in \mathbb{C}$ and $Re(s) >1$?

I would really appreciate any help with this problem.

Answer 1

From the prime number theorem we have $p_n\sim n\log n$, and thus $p_n\lt2n\log n$ for all but finitely many $n$. So

\begin{eqnarray} \sum_{j=1}^k\log p_j &=& O(1)+\sum_{j=1}^k\log\left(2j\log j\right) \\ &=& O(1)+k\log2+\sum_{j=1}^k\log j+\sum_{j=1}^k\log\log j \\ &=& O\left(\sum_{j=1}^k\log j\right) \\ &=& O(k\log k)\;, \end{eqnarray}

and thus

$$ \frac{\sum_{j=1}^k\log p_j}{p_{k+1}^s}\sim\frac{O(k\log k)}{(k\log k)^s}\sim0\;. $$