Manual calculation doesn't match Wolfram Alpha. Why? Say I want to evaluate this sum:
$$\sum_{x=2}^\infty \ln(x^3+1)-\ln(x^3-1)$$
We can rewrite the sum as
$$\ln\left(\prod_{x=2}^\infty \frac{x^3+1}{x^3-1}\right)$$
We can split the product into two products:
$$\ln\left(\prod_{x=2}^\infty \frac{x+1}{x-1}\right)+\ln\left(\prod_{x=2}^\infty \frac{x^2-x+1}{x^2+x+1}\right)$$
These are both telescopic products! We can rewrite them as 
$$\ln\left(\prod_{x=2}^\infty \frac{(x+2)-1}{x-1}\right)+\ln\left(\prod_{x=2}^\infty \frac{(x-1)^2+(x-1)+1}{x^2+x+1}\right)$$
Plugging in numbers makes this pattern more obvious;
$$\ln\left(\prod_{x=2}^\infty \frac{\color{green}{3}}{1}\frac {\color{red}{4}}{2} \frac{\color{blue}{5}}{\color{green}{3}} \frac{\color{bluedark}{6}}{\color{red}{4}} \right) +
\ln\left(\prod_{x=2}^\infty \frac{3}{\color{green}{7}} \frac{\color{green}{7}}{\color{red}{13}} \frac{\color{red}{13}}{\color{blue}{21}} \right)$$
Simplifying, we get that the sum is equal to
$$\ln(1/2)+\ln(3)=\ln\left(\frac 32 \right) \approx 0.405465108108164381978013115464349136571990423462494197614$$
However, when I put the sum in Wolfram Alpha directly, I get the following number: 
$$0.4054588737136331726677370820628648457601892466568342890929$$

Why are these two numbers different? It's not a small difference either; it's on the 5th number after the decimal point! How can Wolfram make such an error? 


When you split the product into two products, the first diverges to $\infty$ and the second goes to zero.  This gives you the indeterminate $\infty-\infty$.  All bets are off.
