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Viefleur [7K]
3 years ago
11

Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

nter your answer using interval notation.) i =
Mathematics
1 answer:
Bingel [31]3 years ago
6 0
Assuming the series is

\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}

The series will converge if

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|

We have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|

So the series will certainly converge if -1, but we also need to check the endpoints of the interval.

If x=1, then the series is a scaled harmonic series, which we know diverges.

On the other hand, if x=-1, by the alternating series test we can show that the series converges, since

\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0

and is strictly decreasing.

So, the interval of convergence for the series is -1\le x.
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