An Euler-proof that cannot be repaired? Ed Sandifer writes on Euler's wonderful work on prime numbers: The sum of the series of reciprocals of prime numbers $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...$ is infinitely large, and is infinitely less than the harmonic series $1 +\frac{1}{2}+\frac{1}{3}+...$. Moreover, the first sum is almost the logarithm of the second sum. Euler’s proof of this again on adding and subtracting series that do not converge, and so does not meet modern standards of rigor. Unfortunately, Euler’s proof can’t really be “repaired”.
http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2029%20infinitely%20many%20primes.pdf
Is it really true that this proof can't be repaired with modern tools? If it can be repaired, however, what has to be done?
 A: The proof is easily repaired (although I'm not sure exactly what the details of the proof you're referring to are and don't know how closely you require the repairing of a proof to adhere to the original proof). Observe that
$$\prod_{p \le n} \left( \frac{1}{1 - \frac{1}{p}} \right) \ge \sum_{k=1}^n \frac{1}{k}.$$
Now take logarithms, use the fact that $\log \frac{1}{1 - x} = x + O(|x|^2)$, and let $n \to \infty$. 
A: This result of Euler is the last theorem in http://eulerarchive.maa.org/docs/originals/E072.pdf, and if you prefer English to Latin look at the last theorem in http://eulerarchive.maa.org/docs/translations/E072en.pdf.  I don't know why Sandifer says the proof can't be repaired. Euler's goal is to show $\sum_{p} 1/p$ diverges, and that in some sense $\sum_{p \leq n} 1/p$ diverges like $\log(\log n)$. (Euler didn't speak directly about asymptotic approximations, and he wrote the last conclusion as $\sum_p 1/p = l(l(\infty))$, using $l$ for the natural logarithm.)
Euler's proof: for each positive integer $k$, set $A_k = \sum_p 1/p^k$.  Then 
$$
e^{\sum_{k \geq 1} A_k/k} = \sum_{n \geq 1} \frac{1}{n}
$$
and $\sum_{k \geq 2} A_k/k < \infty$, so $A_1$, which is $\sum_p 1/p$, diverges. 
Since $A_1 = \infty$ while $\sum_k A_k/k < \infty$, we have 
$$
e^{A_1} = \sum_{n \geq 1} \frac{1}{n}, 
$$
so 
$$
\sum_p \frac{1}{p} = A_1 = \log\left(\sum_{n \geq 1} \frac{1}{n}\right).
$$
QED
The last part is kind of a bogus calculation, but the rest can be fixed using the zeta-function.  


*

*In place of $A_k$ use $\sum_{p} 1/p^{ks}$, so the replacement of the first displayed equation above is 
$$
e^{\sum_{p,k} 1/kp^{ks}} = \sum_{n \geq 1} \frac{1}{n^s}
$$
for ${\rm Re}(s) > 1$.  In other words, this is saying $\sum_{p,k} 1/kp^{ks}$ is a logarithm of $\zeta(s)$ for ${\rm Re}(s) > 1$. Series are absolutely convergent, so all manipulations of terms in the series are justified. Euler does all his work at $s = 1$, where expressions may be divergent.

*Euler's claim that $\sum_{k \geq 2} A_k/k < \infty$ is correct, and the modern replacement for this is the more general observation that $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$. Euler's statement is at $s=1$.

*Let $s$ be real and greater than 1, so the displayed equation can be rewritten as 
$$
\sum_{k \geq 1} \sum_{p} \frac{1}{kp^{ks}} = \log\left(\sum_{n \geq 1} \frac{1}{n^s}\right).
$$
As $s \rightarrow 1^+$, $\sum_n 1/n^s \rightarrow \infty$, so the right side above tends to $\infty$ as $s \rightarrow 1^+$. On the left side above, split off the terms where $k=1$ from the terms where $k \geq 2$.  Since $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$, this series has a limit as $s \rightarrow 1^+$ (Dirichlet series are continuous on open half-planes where they converge). Therefore the sum of the terms with $k = 1$, $\sum_p 1/p^s$, must tend to $\infty$ as $s \rightarrow 1^+$.  For $s > 1$ we have $\sum_p 1/p > \sum_p 1/p^s$, and the right side tends to $\infty$ as $s \rightarrow 1^+$, so $\sum_p 1/p = \infty$.
