Copy from a script I've written a year ago: Every $k \in \mathbb{N}$ can be written as $k=t \cdot s^{2}$ with $t$ not divisible by a square $>1$. This gives the inequality \[\sum_{k=1}^{n}\frac{1}{k}\leq \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \sum_{s=1}^{n}{\frac{1}{s^{2}}}.\] Since $\frac{1}{s^{2}}\leq \frac{1}{s(s-1)}= \frac{1}{s-1}-\frac{1}{s}$ for $s \geq 2$ we get $\sum_{s=1}^{n}{\frac{1}{s^{2}}}\leq 1+\sum_{s=2}^{n}\frac{1}{s-1}-\frac{1}{s}= 2-\frac{1}{n}\leq 2$. Together with the easy to verify property $1+x \leq e^{x}$ for all $x \in \mathbb{R}$ this yields \[\sum_{k=1}^{n}\frac{1}{k}\leq \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \sum_{s=1}^{n}{\frac{1}{s^{2}}}\leq 2 \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \leq 2 \cdot \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}e^\frac{1}{p}= 2 \cdot e^{\sum_{}\frac{1}{p}}\] where the last sum also runs over all primes $p \leq n$. To show that $\sum_{ p \in \mathbb{P}}\frac{1}{p}$ diverges, it now suffices to show that $\sum_{k=1}^\infty \frac{1}{k}$ diverges. But the latter one is a well known property, shown by the following since for all $n \geq 2^{m}$ we have: \[\sum_{k=1}^{n}\frac{1}{k}\geq \sum_{k=1}^{2^{m}-1}\frac{1}{k}= \sum_{s=0}^{m-1}\sum_{k=2^{s}}^{2^{s+1}-1}\frac{1}{k}\geq \sum_{s=0}^{m-1}\sum_{k=2^{s}}^{2^{s+1}-1}\frac{1}{2^{s+1}}= \sum_{s=0}^{m-1}\frac{1}{2}= \frac{m}{2},\] giving the divergence because $m$ can be chosen arbitrary large.

Peter wrote: Let $ p_{n}$ again denote the $ n$th prime number. Show that the infinite series \[ \sum^{\infty}_{n=1}\frac{1}{p_{n}}\] diverges. Reference: A Classical Introduction to Modern Number Theory by K. Ireland and M. Rosen Second Solution We first prepare a lemma. Let $ \varrho(n)$ denote the set of prime divisors of $ n$. Let $ {\mathcal{S}}_{n}(N)$ denote the set of positive integers $ i\leq N$ satisfying that $ \varrho(i)\subset\{ p_{1},\cdots, p_{n}\}$. Lemma. We have $ \left\vert{\mathcal{S}}_{n}(N)\right\vert\leq 2^{n}\sqrt{N}$. Proof It is because every positive integer $ i\in{\mathcal{S}}_{n}(N)$ has a unique factorization $ i = s t^{2}$, where $ s$ is a divisor of $ p_{1}\cdots p_{n}$ and $ t\leq\sqrt{N}$. In other words, $ i\mapsto (s,t)$ is an injective map from $ {\mathcal{S}}_{n}(N)$ to $ \mathcal{T}_{n}(N) =\{\, (s,t)\;\vert\; s\,\vert\, p_{1}\cdots p_{n},\, t\leq\sqrt{N}\}$, which means that \[ \left\vert{\mathcal{S}}_{n}(N)\right\vert\leq\left\vert\mathcal{T}_{n}(N)\right\vert\leq 2^{n}\sqrt{N}.\] Now, assume to the contrary that the infinite series $ \frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots$ converges. Then we can take a sufficiently large positive integer $ n$ satisfying that \[ \frac{1}{2}\geq\sum^{\infty}_{i>n}\frac{1}{p_{i}}=\frac{1}{p_{n+1}}+\frac{1}{p_{n+2}}+\cdots .\] Take a sufficiently large positive integer $ N$ so that $ N > 4^{n+1}$. By its definition of $ {\mathcal{S}}_{n}(N)$, we see that each element $ i$ in $ \{1,\cdots, N\}-{\mathcal{S}}_{n}(N)$ is divisible by at least one prime $ p_{j}$ for some $ j > n$. Since the number of multiples of $ p_{j}$ not exceeding $ N$ is $ \left\lfloor\frac{N}{p_{j}}\right\rfloor$, we have \[ \left\vert\{1,\cdots, N\}-{\mathcal{S}}_{n}(N)\right\vert\leq\sum_{j>n}\left\lfloor\frac{N}{p_{j}}\right\rfloor\] or \[ N-\left\vert{\mathcal{S}}_{n}(N)\right\vert\leq\sum_{j>n}\left\lfloor\frac{N}{p_{j}}\right\rfloor\leq\sum_{j>n}\frac{N}{p_{j}}\leq\frac{N}{2}\] or \[ \frac{N}{2}\leq\left\vert{\mathcal{S}}_{n}(N)\right\vert.\] It follows from this and from \textsc{Lemma 3.2} that \[ \frac{N}{2}\leq 2^{n}\sqrt{N}\] so that $ N\leq 4^{n+1}$. However, this is a contradiction for the choice of $ N$. Third Solution We use an auxiliary inequality without a proof. Lemma. We have $ \frac{1}{1-t}\leq e^{t+t^{2}}$ for all $ t\leq\left[0,\frac{1}{2}\right]$. Let $ l\in\mathbb{N}$. By Fundamental Theorem of Arithmetic, each positive integer $ i\leq p_{l}$ has a unique factorization $ i ={p_{1}}^{e_{1}}\cdots{p_{1}}^{e_{l}}$ for some $ e_{1},\cdots, e_{l}\in\mathbb{Z}_{\geq 0}$. It follows that \[ \sum_{i=1}^{p_{l}}\frac{1}{i}\leq\sum_{ e_{1},\cdots, e_{l}\in\mathbb{Z}_{\geq 0}}\frac{1}{{p_{1}}^{e_{1}}\cdots{p_{l}}^{e_{l}}}=\prod_{j=1}^{l}\left(\sum_{k=0}^{\infty}\frac{1}{{p_{j}}^{k}}\right)=\prod_{j=1}^{l}\frac{1}{ 1-\frac{1}{p_{j}}}\leq\prod_{j=1}^{l}e^{\frac{1}{p_{j}}+\frac{1}{{p_{j}}^{2}}}\] so that \[ \sum_{j=1}^{l}\left(\frac{1}{p_{j}}+\frac{1}{{p_{j}}^{2}}\right)\geq\ln\left(\sum_{i=1}^{p_{l}}\frac{1}{i}\right).\] Together with the estimation \[ \sum_{j=1}^{\infty}\frac{1}{{p_{j}}^{2}}\leq\sum_{j=1}^{l}\frac{1}{(j+1)^{2}}\leq\sum_{j=1}^{l}\frac{1}{(j+1)j}=\sum_{j=1}^{l}\left(\frac{1}{j}-\frac{1}{j+1}\right)=1,\] we conclude that \[ \sum_{j=1}^{l}\frac{1}{p_{j}}\geq\ln\left(\sum_{i=1}^{p_{l}}\frac{1}{i}\right)-1 .\] Since the divergence of the harmonic series $ 1+\frac{1}{2}+\frac{1}{3}+\cdots$ is well-known, now we get the result. Note The fact that the sum of the reciprocals of all prime numbers diverges was first proved by Euler. Using the well-known result $ 1+\frac{1}{2}+\cdots+\frac{1}{N}\geq\ln (N+1)$ and modifying the method in the second solution, we get the following result. For all real numbers $ x\geq 2$, we obtain \[ \sum_{p: prime,\; p\leq x}\frac{1}{p}\geq\ln\left(\ln x\right)-1.\] It turns out that this lower bound inequality is sharp. In fact, it is known that \[ \mathcal{M}=\lim_{n\rightarrow\infty }\left(\sum_{p: prime,\; p\leq n}\frac{1}{p}-\ln\left(\ln n\right)\right)\] converges. Here, $ \mathcal{M}$ is called Meissel-Mertens constant. It is known that $ \mathcal{M}= 0.2614972128476427837554268386086958590516\cdots$

Peter wrote: Let $ p_{n}$ again denote the $ n$th prime number. Show that the infinite series \[ \sum^{\infty}_{n=1}\frac{1}{p_{n}}\] diverges. Reference. Elementary Number Theory by David M. Burton It can be deduced from Prime Number Theorem Let $ \pi(x)$ denote the prime counting function. Since $ pi(x)\rightarrow\frac{ x }{\ln x}$ as $ x\rightarrow\infty$, we can find a constant $ \lambda > 0$ satisfying that \[ \pi(x) >\lambda\frac{ x }{\ln x}\] for all sufficiently large positive real numbers $ x$. This means that $ n >\lambda\frac{ p_{n}}{\ln p_{n}}$ when $ n$ is sufficiently large. Since $ \lambda\frac{ x }{\ln x}>\sqrt{x}$ for all sufficiently large $ x > 0$, we also have $ n >\lambda\frac{ p_{n}}{\ln p_{n}}>\sqrt{ p_{n}}$ or $ n^{2}> p_{n}$ for all sufficiently large $ n$. We conclude that \[ n >\lambda\frac{ p_{n}}{\ln p_{n}}>\lambda\frac{ p_{n}}{\ln\left( n^{2}\right)}\] or \[ \frac{1}{ p_{n}}>\frac{\lambda }{2n\ln n}.\] when $ n$ is sufficiently large. Since \[ \sum_{n=2}^{\infty}\frac{1 }{n\ln n}=\infty\] Comparison Test yields that \[ \sum_{n=1}^{\infty}\frac{1}{ p_{n}}=\infty.\]

Peter wrote: Let $ p_{n}$ again denote the $ n$th prime number. Show that the infinite series \[ \sum^{\infty}_{n=1}\frac{1}{p_{n}}\] diverges. On the other hand, Viggo Brun established that the sum of the reciprocals of the twin primes converges. In other words, \[ \mathcal{B}=\sum_{p, p+2: prime}\left(\frac{1}{p}+\frac{1}{p+2}\right) =\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{7}\right)+\left(\frac{1}{11}+\frac{1}{13}\right)+\cdots\] converges. The constant $ \mathcal{B}=1.90216\cdots$ is called Brun's Constant.

Other solution: Lemma Consider $ a_i\in (0,1)$ Then if $ \prod_{i = 1}^{ + \infty}(1-a_i) = 0$ then $ \sum_{i = 1}^{ + \infty} a_i = + \infty$ Now we has : $ \prod_{k = 1}^{ + \infty}(\sum_{i = 0}^n\frac {1}{p_k^i}) = \sum_{i = 1}^n\frac {1}{i}$ because $ \sum_{k = 1}^{ + \infty}\frac {1}{k} = + \infty$ so $ \prod _{i = 1}^{ + \infty}(1 - \frac {1}{p_i}) = 0$ Imply that $ \sum_{k = 1} \frac {1}{p_k} = + \infty$

How do you prove the Lemma? I only see $ \sum a_i\ge1$ by Bernouilli... And why is $ \prod_{k = 1}^{ + \infty}(\sum_{i = 0}^n\frac {1}{p_k^i}) = \sum_{i = 1}^n\frac {1}{i}$?

And why is $ \prod_{k = 1}^{ + \infty}(\sum_{i = 0}^n\frac {1}{p_k^i}) = \sum_{i = 1}^n\frac {1}{i}$?[/quote] Because every number can represent in the form $ n=\prod_{i=1}^{+\infty}p_i^{r_i}$ Where $ r_i\geq 0$

That means that $ \sum_{i = 0}^n\prod_{k = 1}^{ + \infty}\frac {1}{p_k^i} = \sum_{i = 1}^n\frac {1}{i}$, but how are sum and product commutative here?

$ \prod _{i = 1}^{ + \infty}(1 + \frac {1}{p_1} + ... + \frac {1}{{p_1}^n} = \sum_{i = 1}^{ + \infty}\frac {1}{i}$

Aah, now I get it. Sorry, it seems I'm not very sharp today.