Alternating Prime Sum

Question

How can one evaluate this sum: $\sum_{n=1}^\infty (-1)^{n+1}/{p_n}$? I've tried it for myself but didn't get any satifying answer although it does converge. I also can't find this sum anywhere else and not even Wolfram Alpha knows what to do.

Answer 1

I just found an upper bound function by myself.

We redefine the summation as below;

$$\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{p_{n}} = \sum_{n=1}^{\infty } \frac{p_{2n}-p_{2n-1}}{p_{2n}.p_{2n-1}}$$

According to prime number theorem, $\int_{0}^{x} \frac{dt}{logt} \sim π(x)$. When we take the derivative of both sides we get;

$\frac{1}{log(x)} \sim π'(x)$, where π'(x) represents the derivative of prime counting function according to x. (note that this is only an assumption. Derivatives of two asymptotic functions don't have to be asymptotic again.)

Now, π'(x) is the prime density function around x. For instance, around 100, approximately 1 of every log(100) numbers will be prime. Under this assumption we can say;

$$p_{2n-1} = p_{2n} - log(p_{2n})$$

If we substitute this into the top equation, we get;

$$\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{p_{n}} = \sum_{n=1}^{\infty } \frac{log(p_{2n})}{p_{2n}.(p_{2n} - log(p_{2n}))}$$ Using Riemann-Stieltjes integral method ($\sum_{p primes}^{N} f(p) \sim \int_{2}^{N} \frac{f(x)}{log(x)}dx$), above summation can be represented an integral like below:

$$\sum_{n=1}^{\infty } \frac{log(p_{2n})}{p_{2n}.(p_{2n} - log(p_{2n}))} \sim \frac{1}{2} \int_{2}^{\infty } \frac{dx}{x(x-logx)}$$

Since the indices increase by two, we wrote 1/2 at the beginning of the integral.

Here we can find a better approach if we add the first few terms manually and express the rest by integral;

$$\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{p_{n}} \sim \sum_{n=1}^{x } \frac{(-1)^{n+1}}{p_{n}} + O(\int_{x}^{\infty } \frac{dx}{x(x-logx)})$$

"O" is big O notation here. If we consider the result of this sum as a fixed number k,

$$f(x) = \sum_{n=1}^{x } \frac{(-1)^{n+1}}{p_{n}} + \frac{1}{2} \int_{x}^{\infty } \frac{dt}{t(t-logt)}$$ $$\lim_{x\rightarrow \infty } f(x) = k$$

I found the value of k = 0.269606356970929... by summing the values up to 100000000 by hand (I mean computer). And in the table below we compared the values ​​we obtained using this approach with the exact summation k for different x's.

x	f(x)	$|\frac{f(x)-k}{k}|$
3	0.39737469802526604	0.473907004604176
10	0.28225778161237654	0.046925542793532
96	0.27035676966262771	0.002783364235657
1000	0.26958796380480553	0.000068222301321
10070	0.26960698064666632	0.000002313282759

In order to make a good comparison in the table above, we made sure to write the x values ​​that will return even values ​​when we calculate π(x). That's why it's not all round numbers.

With better π'(x) approximations, better bounds can be found.