Prime Factorial Reciprocals

By Bertrand's postulate , $p_{n+1} < 2p_{n}$ for all natural $n$ , thus $\frac{p_{n+1}}{P_n} < \frac{2p_n}{P_n} = \frac{2}{P_{n-1}}$ .

Next, $P_{n+1} = p_{n+1} \cdot P_n > 2P_n$ . Together with $P_1 = 2$ , we can induct to show that $P_n \ge 2^n$ for all natural $n$ .

Putting them all together,

$\begin{aligned}\displaystyle \sum_{n=1}^\infty \left| \frac{(-1)^{n+1}p_{n+1}}{P_n} \right| &= \sum_{n=1}^\infty \frac{p_{n+1}}{P_n} \\ &= \frac{p_2}{P_1} + \sum_{n=2}^\infty \frac{p_{n+1}}{P_n} \\ &< \frac{3}{2} + \sum_{n=2}^\infty \frac{2}{P_{n-1}} \\ &\le \frac{3}{2} + \sum_{n=2}^\infty \frac{2}{2^{n-1}} \\ &= \frac{3}{2} + \sum_{n=2}^\infty 2^{2-n} \\ &= \frac{3}{2} + 2 = \frac{7}{2} \end{aligned}$

Thus $\sum_{n=1}^\infty \left| \frac{(-1)^{n+1}p_{n+1}}{P_n} \right|$ is bounded above. Additionally, it's clearly monotonically increasing, and thus it converges. This proves that $\sum_{n=1}^\infty \frac{(-1)^{n+1}p_{n+1}}{P_n}$ converges absolutely .

Let $f(n) = \dfrac{(-1)^{n+1} p_{n+1}}{P_{n}}.$ Now by Bertrand's postulate we know that for $n \ge 1$ we have that $p_{n+1} \lt 2p_{n}.$ Thus

$|f(n)| = \dfrac{p_{n+1}}{P_{n}} \lt \dfrac{2p_{n}}{P_{n}} = \dfrac{2}{P_{n-1}}$ for $n \gt 1.$

Also, since $p_{n} \ge 2$ for all $n$ we know that $P_{n-1} \ge 2^{n-1}$ for all $n \gt 1.$ Thus

$|f(n)| \lt \dfrac{2}{P_{n-1}} \le \dfrac{4}{2^{n}}$ for $n \ge 2,$ and so

$\displaystyle\sum_{n=1}^{\infty} |f(n)| = |f(1)| + \displaystyle\sum_{n=2}^{\infty} |f(n)| \lt \dfrac{3}{2} + 4*\sum_{n=2}^{\infty} \left(\dfrac{1}{2}\right)^{n} = \dfrac{3}{2} + 4*\dfrac{\frac{1}{4}}{1 - \frac{1}{2}} = \dfrac{7}{2}.$

As we have now established an upper bound for $\displaystyle\sum_{n=1}^{\infty} |f(n)|$ we can conclude that the given series is absolutely convergent.

(Note that we can establish a tighter upper bound by isolating more of the initial terms. Using a similar argument, if we isolate $|f(1)|$ and $|f(2)|$ we can establish an upper bound of $\frac{8}{3}.$ However, for the purposes of this question the initial bound of $\frac{7}{2}$ will suffice.)

For this problem, we will make use of Bertand's Postulate .

We will also use the fact that where $c$ is a nonzero constant, $\displaystyle\sum_{n=1}^\infty a_n$ and $c\displaystyle\sum_{n=1}^\infty a_n$ have the same convergence properties.

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Bertrand's postulate states that between a number $n$ and the number $2n,$ there exists at least one prime. A direct consequence of this is that between a prime and its double (noninclusive), there is at least one other prime. Thus, $p_n<p_{n+1}<2p_n$ .

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For now, we will ignore the $(-1)^n$ term and determine if $\displaystyle\sum_{n=1}^\infty\dfrac{p_{n+1}}{P_n}$ is divergent. (Obviously, if $\sum a_n$ is absolutely convergent, then the series formed when alternate terms are negated is also convergent.)

Now we can determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\dfrac{p_{n+1}}{P_n}$ .

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Bertrand's postulate says that $p_{n+1}<2p_n,$ so we can state the following. $\sum_{n=1}^\infty\dfrac{p_{n+1}}{P_n}<2\sum_{n=1}^\infty\dfrac{p_n}{P_n}=2\sum_{n=1}^\infty\dfrac{1}{P_{n-1}}$ We can now say that if $2\displaystyle\sum_{n=1}^\infty\dfrac{1}{P_{n-1}}$ is convergent, then $\displaystyle\sum_{n=1}^\infty\dfrac{p_{n+1}}{P_n}$ is convergent as well.

Multiplying a series by a constant does not affect its convergence. We can modify our previous statement to say that if $\displaystyle\sum_{n=1}^\infty\dfrac{1}{P_{n-1}}$ is convergent, then $\displaystyle\sum_{n=1}^\infty\dfrac{p_{n+1}}{P_n}$ is convergent as well.

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Now we will examine the convergence of $\displaystyle\sum_{n=1}^\infty\dfrac{1}{P_{n-1}}$ .

All prime numbers are greater than or equal to $2,$ so we know that $P_k\ge2^k$ for all positive integer $k.$ Thus, $\dfrac{1}{P_k}\le\dfrac{1}{2^k}$ for all positive integer $k.$ Thus, we can say the following.

$\sum_{n=1}^\infty\dfrac{1}{P_{n}}\le\sum_{n=1}^\infty\dfrac{1}{2^n}=1$ By the very large chain of convergies that we have established in this proof, we have that $\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^{n+1}p_{n+1}}{P_n}<1,$ meaning that the series is $\boxed{\text{absolutely convergent}}$