4 years ago

Please Solve this 10+2y

Mathematics

1 answer:

Zigmanuir [339]4 years ago

6 0

I believe the answer is 5y :)

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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

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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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