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?

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 $-1$.
