Exercise 8.10

Question

Exercise 10: 
    Prove that $\sum 1/p$ diverges, the sum extends over all primes.

analambanomenos · Accepted Answer

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Following the hint, let $p_1,\ldots,p_k$ be distinct primes. Then the product of the convergent geometric series of $p_k^{-1}$ yields $$\prod_{j=1}^k\bigg(1+\frac{1}{p_j}+\frac{1}{p_j^2}+\cdots
    \bigg)=\sum\frac{1}{p_1^{n_1}\cdots p_k^{n_k}}$$ where the sum is over all ordered $k$-tuples $(n_1,\ldots,n_k)$ of non-negative integers.

Hence if $N$ is a positive integer, and $p_1,\ldots,p_k$ are the prime numbers that divide at least one integer $\le N$, we have, using Theorem 3.26, $$\sum_{n=1}^N\frac{1}{n}\le\prod_{j=1}^k
    \bigg(1+\frac{1}{p_j}+\frac{1}{p_j^2}+\cdots\bigg)=\prod_{j=1}^k\bigg(1-\frac{1}{p_j}\bigg)^{-1}.$$

Let $f(x)=2x+\log(1-x)$ for $x\in[0,1/2]$. Then $f'(x)=(1-2x)/(1-x)\ge 0$ in this interval. Since $f(0)=0$, we have $f(x)\ge 0$ for $0\le x\le 1/2$, or
    \begin{align*}
        0 &\le 2x+\log(1-x) \
        \log(1-x)^{-1} &\le 2x \
        (1-x)^{-1} &\le e^{2x}.
    \end{align*}
    Applying this to the above, we get 
    \begin{align*}
        \sum_{n=1}^N\frac{1}{n} &\le \prod_{j=1}^k\bigg(1-\frac{1}{p_j}\bigg)^{-1} \
        &\le \prod_{j=1}e^{2/p_j} \
        &= \exp\Bigg(2\sum_{j=1}^k\frac{1}{p_j}\Bigg).
    \end{align*}
    Since the harmonic series diverges, so must the series in the argument of $\exp$ above.

Exercise 8.10

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Exercise 8.10

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