Yes. If the Maclaurin expansion of a function locally converges to the function, then you know the function is smooth. In addition, if the residual of the Maclaurin expansion converges to 0, the function is analytic.
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Best example is that an "odd" (or "even") function's Maclaurin series only has terms with odd (or even) powers. cos(x) and sin(x) are examples of odd and even functions with easy to calculate Maclaurin series.
The numerical value of pi is often found using a Taylor or Maclaurin series (Taylor series centered at 0).
There is no simple answer. There are simple formulae for simple sequences such as arithmetic or geometric progressions; there are less simple solutions arising from Taylor or Maclaurin series. But for the majority of sequences there are no solutions.
There are several uses. For example: * When analyzing curves, the second derivative will tell you whether the curve is convex upwards, or convex downwards. * The Taylor series, or MacLaurin series, lets you calculate the value of a function at any point... or at least, at any point within a given interval. This method uses ALL derivatives of a function, i.e., in principle you must be able to calculate the first derivative, the second derivative, the third derivative, etc.
A Maclaurin series is centered about zero, while a Taylor series is centered about any point c. M(x) = [f(0)/0!] + [f'(0)/1!]x +[f''(0)/2!](x^2) + [f'''(0)/3!](x^3) + . . . for f(x). T(x) = [f(c)/0!] + [f'(c)/1!](x-c) +[f''(c)/2!]((x-c)^2) + [f'''(c)/3!]((x-c)^3) + . . . for f(x).