Is $1/5+1/7+1/11+1/13+1/17+1/19+\cdots$ convergent?

Question

I am struggling to calculate $$\dfrac 15+\dfrac 17+\dfrac 1{11}+\dfrac 1{13}+\dfrac 1{17}+\dfrac 1{19}+\cdots$$

The denominator is 6k - 1 and 6k + 1, k is pos integer from 1 to infinity. Can somebody give me any hint if it is convergent, and if yes, what is the value at infinity? If divergent, why?

As it includes all primes in the demoninator, plus all non primes with 6k - 1 and 6k + 1 should it suggest that it is divergent?

The denominator is 6k - 1 and 6k + 1, k is pos integer from 1 to infinity. — Tilsight, Jul 11 '19 at 08:05
@MartinR Much easier than my approach, since we only need the divergence of the harmonic series. — Peter, Jul 11 '19 at 08:09
Thanks. In addition, at which k becomes the sum greater than 1? — Tilsight, Jul 11 '19 at 08:13
A nice additional question : Can the partial sum ever be an integer ? — Peter, Jul 11 '19 at 08:18
@Tilsight: The asymptotics of the partial sums of the harmonic series are well-known, look up “harmonic numbers.” They grow $\sim \ln (n)$. — Martin R, Jul 11 '19 at 08:27

score 5 · Accepted Answer · answered Jul 11 '19 at 08:12

5

You asked for a hint, so here it is: $$\frac{1}{6k \pm 1} \ge \frac{1}{7k}$$ for all positive integers $k$.

answered Jul 11 '19 at 08:12

Martin R

113,040

score 1 · Answer 2 · answered Jul 11 '19 at 08:07

1

Since among the denominators , there are all the primes greater than $3$ and since the sum of the reciprocals of the primes diverges to $\infty$, this is also the case with the given sum.

answered Jul 11 '19 at 08:07

Peter

84,454

Thanks. In addition, can we see, at which k is the sum greater than 1? – Tilsight Jul 11 '19 at 08:12
1

According to my calculation (with simple brute force), we exceed $1$ with the summand $\frac{1}{65}$ – Peter Jul 11 '19 at 08:16
1

Thanks, Peter,for your help. – Tilsight Jul 11 '19 at 08:36

Is $1/5+1/7+1/11+1/13+1/17+1/19+\cdots$ convergent?

2 Answers2