Paradox: Summation of natural logarithms 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?
 A: $$\sum_{i=1}^{\infty} \ln(i+2)-\ln(i+4)$$
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.
A: 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 $-1$.
A: 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 $\sum\limits_{i=1}^{\infty}\ln(\frac{1}{i})$.
A: 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 $$S_a-S_b=(a-b) \log (n)+\log \left(\frac{\Gamma (b+1)}{\Gamma (a+1)}\right)+\frac{(a-b) (a+b+1)}{2 n}+O\left(\left(\frac{1}{n}\right)^{3/2}\right)$$
