Finding the interval of convergence and the radius of a power series

[Ratio Test =)]

Homework Statement

Find the radius and interval of convergence of the power series:
\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(n+1)\sqrt{n}}

Homework Equations

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The Attempt at a Solution

My soltion:
the ratio test will gives |x^2|=|x|^2
it converges if |x|^2 < 1
i.e. if |x|<1
i.e. if -1<x<1
But the problem here when i substitute x=1 and x=-1 it will give the same series
Is this right?
The resulting series converges by Integral Test.
Radius = 1

 

[Mark44]

x = - 1 doesn't give the same series as x = 1. If x = -1, the series is
\sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{(n+1)\sqrt{n}}
Since 2n - 1 is odd for all integers n, the numerator of the expression in the summation is always -1. If x = 1, the numerator is always +1.

 

[Ratio Test =)]

x = - 1 doesn't give the same series as x = 1. If x = -1, the series is
\sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{(n+1)\sqrt{n}}
Since 2n - 1 is odd for all integers n, the numerator of the expression in the summation is always -1. If x = 1, the numerator is always +1.

Oh...Thanks
But will not affect the convergence right?
since the series which obtained by multiplying a convergent series by a constant is still convergent.
Right?

 

[Mark44]

Right.

 

[Ratio Test =)]

No No Sorry!
there was a typo!
its 2n-2 not 2n-1 in the series =)
Sorry Again.

 

[Dick]

No No Sorry!
there was a typo!
its 2n-2 not 2n-1 in the series =)
Sorry Again.

Ok, then it is the same series at x=1 and x=(-1). So you only have to check one of them. Is it convergent?

 

[Ratio Test =)]

Thanks you.
you saved my life !
It is easy for me.
It converges by integral test.

 

[Dick]

Thanks you.
you saved my life !
It is easy for me.
It converges by integral test.

Sure. Or you could compare it to the convergent p-series 1/n^(3/2).

 

[Ratio Test =)]

Ohhh
My bad
How in Earth I did not notice this :@

 

[Dick]

Ohhh
My bad
How in Earth I did not notice this :@

I didn't say your way was bad. The integral test works fine.