In Mathematica, the input FullSimplify[Sum[Log[x^3+1]-Log[x^3-1],{x,2,Infinity}]] gives Log[3/2], exactly. In WolframAlpha, the issue is that when you request more digits of accuracy, it converts your input into the command NSum[Log[x^3 + 1] - Log[x^3 - 1], {x, 2, Infinity}, WorkingPrecision -> 104] which of course is insufficient working precision for the number of decimal digits that it displays due to the very slow convergence of the sum.
Somewhat ironically, if you enter Product[(x^3+1)/(x^3-1),{x,2,n}] into WolframAlpha, you get exact output: $$\frac{3n(n+1)}{2(n^2+n+1)},$$ which is correct, furthermore upon taking Limit[3n(n+1)/(2(n^2+n+1)), n -> Infinity], you get the correct answer $3/2$. So it isn't as if WolframAlpha cannot compute the original sum symbolically as Mathematica did. It just needs a little extra help, it seems.
Users of WolframAlpha don't always realize that although it is using the same underlying algorithms as Mathematica, it is not the same thing. There are things that one does that the other does not. Precise control of processing of input, for example, is something that the former does not do.
Update. I believe the above response is not entirely sufficient to explain the behavior of WolframAlpha. When I changed the summand to $$\log \left(1 + \frac{2}{x^3-1}\right),$$ WolframAlpha still gives the wrong result when more digits are requested, despite the fact that the generated code NSum[Log[1 + 2/(x^3 - 1)], {x, 2, Infinity}, WorkingPrecision -> 104] yields the correct result. So this points to an internal inconsistency with WolframAlpha that cannot be solely explained by the insufficient precision used in NSum.
To confirm, in Mathematica I input both variants with NSum, namely
NSum[Log[x^3 + 1] - Log[x^3 - 1], {x, 2, Infinity}, WorkingPrecision -> 104]
as well as
NSum[Log[1 + 2/(x^3 - 1)], {x, 2, Infinity}, WorkingPrecision -> 104]
The first input, as expected, generates a warning NIntegrate::ncvb. Also as expected, the second one does not. But if this is the case, then WolframAlpha should not still present the wrong result in the second case when more digits are requested, given that this is the exact code that it generated. When you open up the computable notebook (click the orange cloud) and evaluate the expression, you are basically running a cloud version of Mathematica. Doing this gets the right answer. So I suspect that there is some kind of bug in WolframAlpha that fails to present the correct output.
Best Answer
WolframAlpha is just some natural language input parsing on top of a limited form of Wolfram Language: the programming language used in Mathematica.
If Mathematica were given
{{1,-1},{0,2}}^r, it would not take the $r^{\text{th}}$ power of the $2 \times 2$ matrix in the usual linear algebra sense; it would instead take the $r^{\text{th}}$ power of each entry. You would have to writeMatrixPower[{{1,-1},{0,2}},r]for the desired result.WolframAlpha is a little bit better at guesing what you want: if you write
{{1,-1},{0,2}}^r, it seems to guess that you mean a matrix power. If you write({{1,-1},{0,2}})^(r), that's too confusing for WolframAlpha, and it falls back to the default (and probably wrong) interpretation.So that's the difference between the two expressions. In the first case, WolframAlpha is taking matrix powers and adding them up. In the second case, WolframAlpha is dealing with each entry of the matrix separately.
If you do not want to rely on WolframAlpha guessing what you want, you can use syntax closer to Mathematica syntax to disambiguate. For example, you can ask WolframAlpha to compute
and when you get the output (which is the same as your first result), you can be certain that WolframAlpha is taking the correct kind of matrix power.