I was inspired in Lemma 3.1 of [1] and in the Theorem 4.12 of [2] to ask about a similar statement that shows Lagarias in his paper as Lemma 3.1.


The Lemma from Lagarias's paper is that if $H(n)=\sum\_{i \leq n} \frac{1}{i}$, then for $n\geq 3$ we have $$e^{H(n)}\log H(n) \geq e^{\gamma}n \log \log n,$$ where $\gamma$ is Euler's constant. 


We define for integers $n\geq 1$
$$h\_n=\sum\_{k=1}^n\frac{1}{p\_k},$$
where $p\_k$ denotes the $k$-th prime number. 



> 
> **Question.** I wondered if it is possible to determine a lower bound as a function of $n$ such that 
>  $$\text{lower bound}(n)\leq \exp(h\_n)\log(h\_n)$$
>  holds $\forall n\geq N$ for some suitable integer $N\geq 1$.
> 
> 
> 


I am asking about a statement similar to the lemma due to Lagarias, including a reference to the literature if available.


References:
-----------


[1] Jeffrey C. Lagarias, [*An Elementary Problem Equivalent to the Riemann Hypothesis*](https://www.jstor.org/stable/2695443), The American Mathematical Monthly, Vol. 109, No. 6 (Jun. - Jul., 2002), pp. 534–543. ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=1908008))


[2] Tom M. Apostol, [*Introduction to Analytic Number Theory*](https://link.springer.com/book/10.1007/978-3-662-28579-4), Undergraduate Text in Mathematics Springer (1976). ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=434929))



A similar lemma to a lemma due to Lagarias, for the partial sums of reciprocal of primes
As pointed out in the comments $h\_n\sim \log\log n +M,$ so for $n$ large enough we can write 
$$e^{h\_n}\log h\_n \sim e^M \log n \log(\log \log n+M)\quad(1).$$ 
For $n$ large the right hand side can be approximated by
$$
e^M \log n \left(\log \log\log n+\frac{M}{\log \log n}\right)
$$


where $M$ is the Mertens-Meissel constant. Whether the difference between any of these two RHS's and $\exp\{h\_n\} \log h\_n$ has a definite sign after some large $n$ is probably hard to prove. In fact in the taylor expansion I used in the approximation above there should probably be a $$-\frac{M^2}{(\log\log n)^2}$$
to ensure a lower bound.


But the question is, what is your goal? 


**Edit:** Computations with Magma online calculator give the following for the left hand side (actual) and the right hand side (approx.) of (1)



```
n= 2^ 10: actual 7.1385169507832789173 approx 7.0886111103586373676 
diff 0.049905840424641549657

n= 2^ 15: actual 12.926526293730941052 approx 12.919605358113649214
diff 0.0069209356172918385427

n= 2^ 20: actual 19.115096320844860339 approx 19.113668933136581336
diff 0.0014273877082790029525

```

At $n=2^{21},$ the largest value where the computation terminates [the online calculator has a time limit] the difference is down to $0.001249\cdots$ but still positive. 


