Solve : $\displaystyle\sum_{k=1}^{n} {\sum_{x=1}^{2^k} \bigg\lfloor\bigg(\dfrac{k}{1 + \pi(x)}\bigg)^{1/k}\bigg\rfloor}$?

Basically, I've been doing some problems involving primes and summation of consecutive primes. I know that there is the imprecise equation for this where $p_1$+$p_2$+$p_3$...+$p_n$ is approximately equal to $n^2\cdot \ln(n)$, but that equation, like I said, is only approximate and also doesn't work so great at low values of n. I found it on another question on math.stackexchange.com about adding consecutive primes.

When I tried it on WolframAlpha to get a more precise answer, I simply asked it to do the work of summing the primes from first to various values of n. It basically spat out the given equation, though less simply expressed where I had to generalize the equation a little. (Don't know if I did that right, either. But I tried.)

I got basically:

$n+\displaystyle\sum_{x=1}^{2^1} \bigg\lfloor{\bigg(\dfrac{1}{1 + \pi(x)}\bigg)}^{1/1}\bigg\rfloor+\sum_{x=1}^{2^2} \bigg\lfloor\bigg(\dfrac{2}{1 + \pi(x)}\bigg)^{1/2}\bigg\rfloor+\sum_{x=1}^{2^3} \bigg\lfloor\bigg(\dfrac{3}{1 + \pi(x)}\bigg)^{1/3}\bigg\rfloor\cdots+ \sum_{x=1}^{2^n} \bigg\lfloor\bigg(\dfrac{n}{1 + \pi(x)}\bigg)^{1/n}\bigg\rfloor$.

So I generalized it how I thought I probably should as a partial sum of the partial sums except excluding the first n. That got the above equation.

So, yeah... Is there anyway you'd know of to solve/simplify the equation I got from WolframAlpha? I don't know enough about primes, floors and sums to do it. And it was too hard for WolframAlpha to do it and not charge me money.

Thanks for any help and for all the explanations I'm bound to receive about how all this works. I'm so lost.

• Formatting tips here. – Em. Apr 16 '17 at 7:33
• WA answered to what ? And $p_n \sim n \ln n$ by the PNT – reuns Apr 18 '17 at 4:11
• So? Is the answer that $n^2 ln(n)$ thing? Because that doesn't seem right. – A. Nutt Apr 23 '17 at 20:10