Have fun with series (2020)

Series (a) diverges; the others converge.

(a) is a pretty well known divergent series; it even has its own Wikipedia page , which lists several proofs.

(b) is absolutely convergent : the sum of the absolute values of the terms in $B$ is

$\sum_{n=1}^{\infty} \left| \frac{1}{p_n^{3/2}}\right| < \sum_{n=1}^{\infty} \left| \frac{1}{n^{3/2}}\right| = \zeta \left( \frac32 \right) \approx 2.61$

where $p_n$ is the $n^{th}$ prime number and $\zeta$ is the Reimann zeta function.

(c) also converges. The pattern is a bit easier to spot if we rewrite it as $S=\frac12+\frac13-\frac24+\frac15+\frac16-\frac27+\cdots$ . We have

$\begin{aligned}S &= \sum_{n=1}^{\infty} \frac{1}{3n-1}+\frac{1}{3n}-\frac{2}{3n+1} \\ &= \sum_{n=1}^{\infty} \frac{9n-1}{27n^3-3n} \quad \text{\blue{note all terms are positive}} \\ &<\sum_{n=1}^{\infty} \frac{9n}{27n^3-3n} \\ &=\sum_{n=1}^{\infty} \frac{3}{9n^2-1} \\ &<\sum_{n=1}^{\infty} \frac{3}{8n^2} \end{aligned}$

which converges to $\frac38 \zeta(2)=\frac{\pi^2}{16}\approx 0.6$ ; hence series (c) also converges. (NB: I've taken very lazy bounds here - but this is still enough to prove convergence)

The prime zeta function is defined as:

$P(s) = \sum_{n > 0} \frac 1{p(n)}$

where $p(n)$ is the $n$ th prime number. $P(s)$ converges if $\Re(s) > 1$ . Since $A = P(1)$ , series $A$ diverges . It is proven that the sum of all primes, which is series $A$ , diverges (see proof ).

Series $B$ can be written as

$B = - \sum_{n >0} \frac 1{p(n)^\frac 32} + 2 \sum_{n >0} \frac 1{p\left(\frac {n(n+1)}2 \right)^\frac 32} = - P\left(\frac 32\right) + 2 Q$

Since, other than the first term, series $Q < P\left(\frac 32\right)$ monotonically and $P\left(\frac 32\right)$ converges, $Q$ converges and therefore series $B$ converges.

Series $C$ converges:

$\begin{aligned} C & = \frac 12 + \frac 13 \blue{- \frac 24} + \frac 15 + \frac 16 \blue{- \frac 27} + \frac 18 + \frac 19 \blue{- \frac 2{10}} + \cdots \\ & = \frac 12 + \frac 13 \blue{- \frac 14 - \frac 14} + \frac 15 + \frac 16 \blue{- \frac 17 -\frac 17} + \frac 18 + \frac 19 \blue{- \frac 1{10} - \frac 1{10}} + \cdots \\ & = \frac 12 + \frac 1{12} - \frac 1{20} + \frac 1{42} - \frac 1{56} + \frac 1{90} - \frac 1{110} + \cdots \end{aligned}$

Series $C$ is an alternating series with $|a_n|$ decreases monotonically and $\displaystyle \lim_{n \to \infty} a_n = 0$ , by alternating series test , series $C$ converges.

Therefore $\boxed 2$ of the series converge.

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Bonus : Finding $C$

$\begin{aligned} C & = \frac 12 + \frac 13 - \frac 24 + \frac 15 + \frac 16 - \frac 27 + \frac 18 + \frac 19 - \frac 2{10} + \cdots \\ & = 2 - \left(2 - \frac 12 - \frac 13 + \frac 24 - \frac 15 - \frac 16 + \frac 27 - \frac 18 - \frac 19 + \frac 2{10} - \cdots\right) \\ & = 2 - 2 \left(\cos 0 + \frac {\cos \frac 23\pi}2 + \frac {\cos \frac 43\pi}3 + \frac {\cos 2\pi}4 + \frac {\cos \frac 83\pi}5 + \cdots\right) \\ & = 2 - 2 \Re \left(1 + \frac {e^{\frac 23\pi i}}2 + \frac {e^{\frac 43\pi i}}3 + \frac {e^{2\pi i}}4 + \frac {e^{\frac 83\pi i}}5 + \cdots\right) \\ & = 2 + 2 \Re \left(\frac {\ln (1-e^{\frac 23 \pi i})}{e^{\frac 23 \pi i}} \right) = 2 + 2 \Re \left(\frac {\ln \left(\frac 32-\frac {\sqrt 3}2i \right)}{-\frac 12 + \frac {\sqrt 3}2i} \right) \\ & = 2 + 2 \Re \left(\frac {\ln \left(\sqrt 3 e^{-\frac \pi 6 i}\right)}{-\frac 12 + \frac {\sqrt 3}2i} \right) = 2 + 2 \Re \left(\frac {3\ln 3 - \pi i}{-3 + 3\sqrt 3 i} \right) \\ & = 2 - \frac 23 \Re \left(\frac {(3\ln 3 - \pi i)(1+\sqrt 3 i}{1+3} \right) \\ & = 2 - \frac {\ln 3}2 - \frac {\sqrt 3 \pi}6 \end{aligned}$