How do you find the interval of convergence #Sigma n3^nx^n# from #n=[0,oo)#?
The series
is convergent for
Given the series:
We then evaluate:
So we have that:
Now if we can consider the partial sums of even order, we have:
while for the partial sums of odd order:
so the series is irregular.
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To find the interval of convergence for the series ( \sum_{n=0}^{\infty} n3^n x^n ), we use the ratio test.

Apply the ratio test: [ \lim_{n \to \infty} \left \frac{a_{n+1}}{a_n} \right = \lim_{n \to \infty} \left \frac{(n+1)3^{n+1} x^{n+1}}{n3^n x^n} \right ]

Simplify the expression: [ = \lim_{n \to \infty} \left \frac{(n+1)3^{n+1}}{n3^n} \right x ] [ = \lim_{n \to \infty} \left \frac{(n+1)}{n} \cdot 3 \right x ] [ = \lim_{n \to \infty} 3 x ]

Apply the limit: [ \lim_{n \to \infty} 3 x = 3 x ]

The series converges if ( 3x < 1 ): [ 3x < 1 \Rightarrow x < \frac{1}{3} ]

Determine the interval of convergence: The series converges absolutely if ( x < \frac{1}{3} ).
Therefore, the interval of convergence for the series ( \sum_{n=0}^{\infty} n3^n x^n ) is ( \left(\frac{1}{3}, \frac{1}{3}\right) ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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