Power series and ratio test: confused about the interval of convergenceI have a question for you. I was asked to find the Maclaurin series of ln ⁡ ( sin ⁡ x / x ) and to evaluate its convergence. After finding the power series, I've applied the ratio test and I've found that the series converges for | − x 2 / 6 + x 4 / 120 | &lt; 1. When I solve the system of inequalities, I find that it is actually impossible, because the first inequality is verified for every real value of x, while the second one has no solution. How can it be? Where do I go wrong?

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Power series and ratio test: confused about the interval of convergenceI have a question for you. I was asked to find the Maclaurin series of   ln  ⁡  (  sin  ⁡  x  /  x  ) and to evaluate its convergence. After finding the power series, I&#039;ve applied the ratio test and I&#039;ve found that the series converges for   |  −      x    2    /  6  +      x    4    /  120  |  &amp;lt;  1. When I solve the system of inequalities, I find that it is actually impossible, because the first inequality is verified for every real value of x, while the second one has no solution. How can it be? Where do I go wrong?

Blaine Foster · Accepted Answer

The firsts terms of the function   log  ⁡  (  sin  ⁡  x  /  x  ) are   −      1    6        x    2    −      1    180        x    4    +  ⋯, so you may want to check your calculations.That mistake apart, knowing only a finite numbers of terms of the series you can&#039;t determine the radius of convergence. To apply the ratio test, for example, you need to calculate       lim          n      →      ∞                  a              n        +        1                    a      n      , so it&#039;s of no use to know that       a    2    =  −      1    6   and       a    4    =  −      1    180  .You&#039;ll need some other way to determine the convergence radius.

Power series and ratio test: confused about the interval of convergence I have a question for you.

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