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The terms in your sum tend to $1$, therefore your sum is asymptotically $k$ (as $k$ tends to infinity), which is $o(p_k)$. Hence your initial condition holds for $k$ sufficiently large, even if you replace $(1+\varepsilon)/e$ by any positive constant (e.g. by $1/100$).

It is also-well known that for given $\varepsilon>0$ and for $n$ sufficiently large (in terms of $\varepsilon$), there is always a prime number between $n$ and $(1+\varepsilon)n$. In fact it is known that, for $n$ sufficently large, there is always a prime number between $n$ and $n+n^{0.525}$.

Added. The OP changed his initial condition from $\sum_{p \le p_k} p^{\frac{1}{p-1}} \lt \frac{1 + ε}{e} p_k$ to $\prod_{p \le p_k} p^{\frac{1}{p-1}} \lt \frac{1 + ε}{e} p_k$. The new condition is false for $\epsilon=1/2$ and $k$ sufficiently large. Indeed, we have $$\prod_{p \le p_k} p^{\frac{1}{p-1}}=\exp\left(\sum_{p \le p_k}\frac{\log p}{p-1}\right)=\exp\left(o(1)+\sum_{n \le p_k}\frac{\Lambda(n)}{n}\right).$$ The right hand side is asymptotically $e^{-\gamma}p_k$, where $\gamma$ is Euler's constant, so the left hand side exceeds $\frac{1.5}{e}p_k$ for $k$ sufficiently large.

The terms in your sum tend to $1$, therefore your sum is asymptotically $k$ (as $k$ tends to infinity), which is $o(p_k)$. Hence your initial condition holds for $k$ sufficiently large, even if you replace $(1+\varepsilon)/e$ by any positive constant (e.g. by $1/100$).

It is also-well known that for given $\varepsilon>0$ and for $n$ sufficiently large (in terms of $\varepsilon$), there is always a prime number between $n$ and $(1+\varepsilon)n$. In fact it is known that, for $n$ sufficently large, there is always a prime number between $n$ and $n+n^{0.525}$.

Added. The OP changed the initial condition from $\sum_{p \le p_k} p^{\frac{1}{p-1}} \lt \frac{1 + ε}{e} p_k$ to $\prod_{p \le p_k} p^{\frac{1}{p-1}} \lt \frac{1 + ε}{e} p_k$. The new condition is false for $\epsilon=1/2$ and $k$ sufficiently large. Indeed, we have $$\prod_{p \le p_k} p^{\frac{1}{p-1}}=\exp\left(\sum_{p \le p_k}\frac{\log p}{p-1}\right)=\exp\left(o(1)+\sum_{n \le p_k}\frac{\Lambda(n)}{n}\right).$$ The right hand side is asymptotically $e^{-\gamma}p_k$, where $\gamma$ is Euler's constant, so the left hand side exceeds $\frac{1.5}{e}p_k$ for $k$ sufficiently large.

The terms in your sum tend to $1$, therefore your sum is asymptotically $k$ (as $k$ tends to infinity), which is $o(p_k)$. Hence your initial condition holds for $k$ sufficiently large, even if you replace $(1+\varepsilon)/e$ by any positive constant (e.g. by $1/100$).

It is also-well known that for given $\varepsilon>0$ and for $n$ sufficiently large (in terms of $\varepsilon$), there is always a prime number between $n$ and $(1+\varepsilon)n$. In fact it is known that, for $n$ sufficently large, there is always a prime number between $n$ and $n+n^{0.525}$.

The terms in your sum tend to $1$, therefore your sum is asymptotically $k$ (as $k$ tends to infinity), which is $o(p_k)$. Hence your initial condition holds for $k$ sufficiently large, even if you replace $(1+\varepsilon)/e$ by any positive constant (e.g. by $1/100$).

It is also-well known that for given $\varepsilon>0$ and for $n$ sufficiently large (in terms of $\varepsilon$), there is always a prime number between $n$ and $(1+\varepsilon)n$. In fact it is known that, for $n$ sufficently large, there is always a prime number between $n$ and $n+n^{0.525}$.

Added. The OP changed his initial condition from $\sum_{p \le p_k} p^{\frac{1}{p-1}} \lt \frac{1 + ε}{e} p_k$ to $\prod_{p \le p_k} p^{\frac{1}{p-1}} \lt \frac{1 + ε}{e} p_k$. The new condition is false for $\epsilon=1/2$ and $k$ sufficiently large. Indeed, we have $$\prod_{p \le p_k} p^{\frac{1}{p-1}}=\exp\left(\sum_{p \le p_k}\frac{\log p}{p-1}\right)=\exp\left(o(1)+\sum_{n \le p_k}\frac{\Lambda(n)}{n}\right).$$ The right hand side is asymptotically $e^{-\gamma}p_k$, where $\gamma$ is Euler's constant, so the left hand side exceeds $\frac{1.5}{e}p_k$ for $k$ sufficiently large.

The terms in your sum tend to $1$, therefore your sum is asymptotically $k$ (as $k$ tends to infinity), which is $o(p_k)$. Hence your initial condition holds for $k$ sufficiently large, even if you replace $(1+\varepsilon)/e$ by any positive constant (e.g. by $1/100$).

It is also-well known that for given $\varepsilon>0$ and for $n$ sufficiently large (in terms of $\varepsilon$), there is always a prime number between $n$ and $(1+\varepsilon)n$. In fact it is known that, for $n$ sufficently large, there is always a prime number between $n$ and $n+n^{0.525}$.

The terms in your sum tend to $1$, therefore your sum is asymptotically $k$ (as $k$ tends to infinity), which is $o(p_k)$. Hence your initial condition holds for $k$ sufficiently large, even if you replace $(1+\varepsilon)/e$ by any positive constant (e.g. by $1/100$).

It is also-well known that for given $\varepsilon>0$ and for $n$ sufficiently large (in terms of $\varepsilon$), there is always a prime number between $n$ and $(1+\varepsilon)n$. In fact it is known that, for $n$ sufficently large, there is always a prime number between $n$ and $n+n^{0.525}$.