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Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 1$, where $N_k$ denotes the $k$-th primorial number $N_k=p_1p_2\cdots p_k$ and $p_j$ the $j$-th prime number?

What I think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?

Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 2$, where $N_k$ denotes the $k$-th primorial number $N_k=p_1p_2\cdots p_k$ and $p_j$ the $j$-th prime number?

What I think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?

Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 1$, where $N_k$ denotes the $k-th$ primorial number and $p_k$ the $k-th$ prime ?

What i think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?

Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 1$, where $N_k$ denotes the $k$-th primorial number $N_k=p_1p_2\cdots p_k$ and $p_j$ the $j$-th prime number?

What I think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?

Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 1$, where $N_k$ denotes the $k-th$ primorial number and $p_k$ the $k-th$ prime ?

What i think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if the inequality holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?

Does the inequality

$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$

hold for all integers $k\geq 1$, where $N_k$ denotes the $k-th$ primorial number and $p_k$ the $k-th$ prime ?

What i think:

By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$

and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?