# On a sum involving prime numbers

I find myself needing the asymtotics of the following summation for my work. Let $a$ be a positive real number and $p_n$ be the $n$-th prime.

$$\sum_{k=1}^{n} [k^a - (k-1)^a]p_k$$

At $a=1$, this becomes the sum of the first $n$ primes and the asymptotics of this is well known. Moreover it is easy to prove that

$$\sum_{k=1}^{n} [k^a - (k-1)^a]p_k = n^a p_n - \int_{2}^{p_n} \pi(x)^a dx.$$

I want an asymptotic expansion in terms of either $n$ or $p_n$ or a combination of both and get rid of the integration. (Don't ask me how many terms of the asymptotic expansion you want, do your best.)

-
When you say "asymptotics", how much of the asymptotics do you mean? Asymptotics for both $p_n$ and $\pi(x)$ are known, but the error terms are not so great (and are not so great even conditionally on GRH) –  Igor Rivin Jun 19 '12 at 14:56
The leading term will be $(a/(a+1))n^{a+1}/\log n$, but the question is how much accuracy you need... –  Anthony Quas Jun 19 '12 at 15:47
@Anthony, There is no limit to accuracy, the more accurate the better. Having said that, there is always a trade off between accuracy and the amount of effort required to obtain that level of accuracy. Unless it is possible to find the general term of the asymptotic expansion of the above sum, I will be happy with an asymptotic expansion that give the first two or three terms followed by an error term. –  Nilotpal Sinha Jun 20 '12 at 6:08

Nature wants to count the primes up to some cutoff point $x$; when we humans insist on labeling the $n$th prime as the $n$th prime, we are destined to have very large error terms. Here, I don't know that you're going to do much better than just substituting in an asymptotic expression for the $k$th prime, changing the problem immediately to something like $$\sum_{k=1}^n [k^a-(k-1)^a] \bigg( k\log k + k\log\log k - k + \frac{k\log\log k - 2k}{\log k} + O\bigg( \frac{k(\log\log k)^2}{\log^2k} \bigg) \bigg),$$ and then estimating each piece of this sum using regular analysis, divorced from number theory.
You can rewrite the sum using prime gap notation. With $d_k=p_{k+1}-p_k$, the sum becomes $$n^ap_n - \sum_{k=1}^{n-1} k^ad_k$$ and now you can use some knowledge of prime gaps to understand the last sum. For purposes of exposition I will ignore the error introduced by pretending $d_1$ is 2 even though it is actually 1. With this pretense, I can call all of the $d_k$ even numbers and with high probability assume they range from 2 to some small even number which conjecturally is at most $(\log n)^2$ but potentially at least $\log {p_n} \log{\log{p_n}}$ : let's call it Fred. I can then break up the sum into 1/2 Fred-many sums of the form $2\sum_{k \in A_i}k^a$. I will let you come up with a careful definition of $A_i$, but $A_1$ should be all the integers between 0 and n since my pretense is that all the $d_k$ are at least 2, $A_2$ will be like $A _1$ but will omit those k for which $d_k$ is exactly 2 and so on. The first sum of the 1/2 Fred-many sums is the largest and is readily computed; cf Bernoulli sums, you should get something of order $n^{a+1}$. The remaining terms get successively smaller until the sum corresponding to the maximal prime gaps is reached. You may find this perspective handy for your work, unless you derived your sum from this kind of expression, in which case, Oops.