Consider the expression : $$\sum_{i=1}^{\infty}\ln(i+2)-\ln(i+4)$$
If one evaluates it out, one gets $$\ln(\frac{3\times4\times5\times6\times...}{5\times6\times7\times8\times...})=\ln(12)$$ That value is positive.
However, each individual term of the original expression is negative, so the sum should be negative.
What's going on there?
On
The problem here is the phrase "if one evaluates it out." The series $\sum\limits_{i=1}^{\infty} \ln(i+2)-\ln(i+4)$ diverges, as we can see by comparing it to the series $\sum\limits_{i=1}^{\infty}\ln(\frac{1}{i})$.
On
In fact, if we ever truncate the sum, we do get a negative number. The first few partial sums are thus
\begin{align} \ln{\frac{3\cdot4}{5\cdot6}}&=\ln{\frac{12}{30}}\\ \\ \ln{\frac{3\cdot4\cdot5}{5\cdot6\cdot7}}&=\ln{\frac{12}{42}}\\ \\ \ln{\frac{3\cdot4\cdot5\cdot 6}{5\cdot6\cdot7\cdot 8}}&=\ln{\frac{12}{56}}\\ \\ \end{align}
What you have done is counted extra terms which contribute to the numerator, but not the terms which contribute to the denominator.
By the same logic, $\sum_{k=1}^\infty k-(k+1)=1-2+2-3+3-4+\cdots=1$. But clearly this sum diverges since you are always adding $-1$.
On
Probably too complex as an answer but too long for a comment.
Let us consider $$S_a=\sum_{i=1}^{n}\ln(i+a)$$ Using Pochhamer notation, it writes $$S_a=\log \left((a+1)_n\right)$$ Using instead the gamma function, it writes $$S_a=\log \left(\frac{\Gamma (a+n+1)}{\Gamma (a+1)}\right)$$ Now, using Stirling approximation for the gamma function when $n$ is large $$S_a=n (\log (n)-1)+\left(\left(a+\frac{1}{2}\right) \log (n)+\log \left(\frac{\sqrt{2 \pi }}{\Gamma (a+1)}\right)\right)+\frac{6 a^2+6 a+1}{12 n}+O\left(\left(\frac{1}{n}\right)^{3/2}\right)$$ which makes $$S_a-S_b=(a-b) \log (n)+\log \left(\frac{\Gamma (b+1)}{\Gamma (a+1)}\right)+\frac{(a-b) (a+b+1)}{2 n}+O\left(\left(\frac{1}{n}\right)^{3/2}\right)$$
$$\sum_{i=1}^{\infty} \ln(i+2)-\ln(i+4)$$ is a Telescoping serie in fact you can highlight the terms that cancel out by expanding the serie $$\sum_{i=1}^{\infty} \ln(i+2)-\ln(i+4) = \ln(3)-\ln(5)+\ln(4)-\ln(6)+\ln(5)-\ln(7)+\ldots \\ = \ln(3)+\ln(4) - \infty = \ln(12) - \infty = -\infty$$ so actually there is not any paradox.