- #1
Denis99
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Find Radius and Interval of Convergence for \(\displaystyle \sum_{1}^{\infty}(\frac{x}{sinn})^{n}\).
I don`t have any ideas how to do that :/
I don`t have any ideas how to do that :/
Country Boy said:The standard method of determining the radius of convergence of power series is to use the "ratio test": [tex]\left|\left(\frac{x^{n+1}}{sin^{n+1}(n+1)}\right)\left(\frac{sin^n(n)}{x^n}\right)\right|= |x|\frac{sin^n(n)}{sin^{n+1}(n+1)}[/tex].
The radius of convergence is 1 over the limit of the fraction.
The ratio test won't work in this example. I think you will find that in this case the radius of convergence is zero. To see that, use the "$n$th term test": if the $n$th term of a series does not tend to zero then the series does not converge.Denis99 said:Find Radius and Interval of Convergence for \(\displaystyle \sum_{1}^{\infty}(\frac{x}{sinn})^{n}\).
I don't have any ideas how to do that :/
The radius of convergence for (x/sin(n))^n is 1, meaning that the series will converge for all values of x within a distance of 1 from the center point 0.
No, the radius of convergence must be a positive value. A negative radius would not make sense in the context of a series.
The radius of convergence is determined by the ratio test, where the limit of the absolute value of the ratio of consecutive terms is taken as n approaches infinity. If the limit is less than 1, the series will converge, and if it is greater than 1, the series will diverge.
The interval of convergence for (x/sin(n))^n is (-1,1), meaning that the series will converge for all values of x between -1 and 1, including the endpoints.
No, the interval of convergence for (x/sin(n))^n will always be (-1,1). This is because the series will diverge for any values of x outside of this interval.