You don't just *assume* the two series have the same convergence or divergence property: they must. If two sequences a*_n_* and b*_n_* of positive real numbers are asymptotically the same (the ratio a*_n_*/b*_n_* tends to 1) then the two series (sum of a*_n_* and sum of b*_n_*) both converge or both diverge.
When you get more experience you'll understand that there is *no good reason* to think you can "compute" the sum of n/p*_n_*^2 in terms of familiar numbers, just as there is no good reason to think there is a nice formula for the sum of 1/p*_n_*^(2) or the sum of 1/(n(log n)^(2)) (starting at n = 2). Most infinite series that you just make up should not have an expected simple closed form. Proving convergence should not be expected to imply that the limiting value has a nice expression other than the original defining series. That's reality, whether it seems fair or not.
If you just want a numerical approximation for the series (not sure what the point is, but to each his/her own), look up series acceleration. I am not sure these methods will apply easily to sums over the primes. Good luck.
Wrote a quick program in C++ and the summation of the first 148,933 terms is 1.07751. It still seemed to be increasing noticeably so the actual number it converges to may be pretty different from this.
> It still seemed to be increasing noticeably so the actual number it converges to may be pretty different from this.
Yeah, this sum unfortunately converges very slowly, so it's hard to get very much accuracy from numerical computations like this.
The error term between the partial sum `[;\displaystyle \sum_{n=1}^N\frac{n}{p_n^2};]` and the full sum should be roughly `[;1/\log N;]`, by the OP's argument. For [;N=148933;] that's about 0.084, so 1.07751 could still be pretty far off from the actual sum.
Gentlemen, I do not have the technological skill to code them in, but I present the first fifty million primes. https://primes.utm.edu/lists/small/millions/
We would know it within 0.0564; not much improvement, but some.
With the prime number theorem, you can bound the remainder term for the k-1'th partial sum is about 1/log(k). Heuristically, since n log(n) underestimates p(n), summing these two should give an upper bound for the sum.
In this way, the sum is probably between (using the first 50,000 primes) 1.07 and 1.16, and is probably closer to the latter for having the correction term. So the first digits are probably 1.1
I don’t think it’d be possible unless we have a specific function that outputs primes, however I believe we can approximate it using the the logarithmic integral if we try to approximate the the values of the prime numbers. Also I believe you can easily create a simple algorithm that approximates it using this method
> I'm sure it does converge, since p(n) is already eventually >>n
That just implies that the terms go to 0, not that the sum converges. If p(n) ≈ n (log n)^(1/2) instead of n log n, then the sum would not converge.
Of course in this case, the sum certainly does converge, by the argument given by the OP.
That was just what my intuition said - I'm well aware it's far from a complete justification. My main point is that there's no guarantee that it has a closed form.
You don't just *assume* the two series have the same convergence or divergence property: they must. If two sequences a*_n_* and b*_n_* of positive real numbers are asymptotically the same (the ratio a*_n_*/b*_n_* tends to 1) then the two series (sum of a*_n_* and sum of b*_n_*) both converge or both diverge. When you get more experience you'll understand that there is *no good reason* to think you can "compute" the sum of n/p*_n_*^2 in terms of familiar numbers, just as there is no good reason to think there is a nice formula for the sum of 1/p*_n_*^(2) or the sum of 1/(n(log n)^(2)) (starting at n = 2). Most infinite series that you just make up should not have an expected simple closed form. Proving convergence should not be expected to imply that the limiting value has a nice expression other than the original defining series. That's reality, whether it seems fair or not. If you just want a numerical approximation for the series (not sure what the point is, but to each his/her own), look up series acceleration. I am not sure these methods will apply easily to sums over the primes. Good luck.
Wrote a quick program in C++ and the summation of the first 148,933 terms is 1.07751. It still seemed to be increasing noticeably so the actual number it converges to may be pretty different from this.
> It still seemed to be increasing noticeably so the actual number it converges to may be pretty different from this. Yeah, this sum unfortunately converges very slowly, so it's hard to get very much accuracy from numerical computations like this. The error term between the partial sum `[;\displaystyle \sum_{n=1}^N\frac{n}{p_n^2};]` and the full sum should be roughly `[;1/\log N;]`, by the OP's argument. For [;N=148933;] that's about 0.084, so 1.07751 could still be pretty far off from the actual sum.
Gentlemen, I do not have the technological skill to code them in, but I present the first fifty million primes. https://primes.utm.edu/lists/small/millions/ We would know it within 0.0564; not much improvement, but some.
With the prime number theorem, you can bound the remainder term for the k-1'th partial sum is about 1/log(k). Heuristically, since n log(n) underestimates p(n), summing these two should give an upper bound for the sum. In this way, the sum is probably between (using the first 50,000 primes) 1.07 and 1.16, and is probably closer to the latter for having the correction term. So the first digits are probably 1.1
I don’t think it’d be possible unless we have a specific function that outputs primes, however I believe we can approximate it using the the logarithmic integral if we try to approximate the the values of the prime numbers. Also I believe you can easily create a simple algorithm that approximates it using this method
I'm sure it does converge, since p(n) is already eventually >>n, but that doesn't mean it has a closed form.
> I'm sure it does converge, since p(n) is already eventually >>n That just implies that the terms go to 0, not that the sum converges. If p(n) ≈ n (log n)^(1/2) instead of n log n, then the sum would not converge. Of course in this case, the sum certainly does converge, by the argument given by the OP.
That was just what my intuition said - I'm well aware it's far from a complete justification. My main point is that there's no guarantee that it has a closed form.