Does this series converge or diverge?

Question

We know the series whose terms are the inverses of the primes diverges.

Now, let p_n be the prime of order 2n -1, that is
p_1 = 2
p_2 = 5
p_3 = 7,
p_4 = 11, and so on

Then, does Sum (n=1, oo) 1/(p_n) converge or diverge?

Thank you

Oh, sorry, there's actually a mistake It is
p_1 = 2
p_2 = 5
p_3 = 11
p_4 = 13, etc

Oh, p_4 = 17, not 13

Answer 1

I think that you may have made a mistake in your problem statement, but I think I can guess what you meant.

Let p_n be the (2n -1)th prime; that is
p_1 = 2
p_2 = 5
p_3 = 11
p_4 = 17, and so on.

I will show that the sum of 1/p_n diverges.

To see this, let q_n be the (2n)th prime; that is
q_1 = 3
q_2 = 7
q_3 = 13
q_4 = 19, and so on.

Now, as you have stated, the series whose terms are the inverse of the primes diverges. This means that either the sum of 1/p_n or the sum of 1/q_n diverges; for if not, we would have the sum of 1/p_n + 1/q_n convergent, a contradiction.

Now, either the sum of 1/p_n diverges or the sum of 1/q_n diverges.

If the sum of 1/p_n diverges, then we are done.

If the sum of 1/q_n diverges, then notice that 1/p_n > 1/q_n for all n. Hence the sum of 1/p_n diverges as well.

Thus, in either case, the sum of 1/p_n is a divergent series.

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It turns out that the sum of 1/q_n is a divergent series as well; just throw away the first term and apply the same argument.

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Relatively straightforward generalization: If {a_n} is a positive, decreasing sequence, and the sum of a_n diverges, then the sum of the odd-indexed terms of a_n diverges, and also the sum of the even-indexed terms of a_n diverges.

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More generalization: If {a_n} is a positive, decreasing sequence, and the sum of a_n diverges, then the sum of the terms with indices congruent to a (mod b) diverges, where a and b are any positive integers, b > 1.

Answer 2

I have to agree with ironduke above, you've generated your numbers incorrectly.

All positive prime numbers are of the form 2n-1 for some n in the naturals except for 2.

To see why, take note that if k is an integer than k is either odd or even. If k is even it is not prime (except for 2). Hence k=2n-1 for some n.

So as your original post says it is true that the series of inverse primes diverges.

I don't see where in lies your question.

Answer 3

Not clear how you are generating your sequence from 2n-1.

For p_1, n = .5
For p_2, n = 3
For p_3, n = 4
For p_4, n= 6