# How do you find the interval of convergence #Sigma (2^nx^n)/(lnn)# from #n=[2,oo)#?

The series:

is convergent for

Given:

Evaluate the ratio:

The limit is then:

which can be proved to be convergent based on Leibniz' theorem, as:

and

In conclusion the series:

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To find the interval of convergence for the series ( \sum_{n=2}^{\infty} \frac{2^n x^n}{\ln n} ):

- Apply the ratio test to determine the convergence interval.
- Apply the ratio test by evaluating ( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ), where ( a_n = \frac{2^n x^n}{\ln n} ).
- If the limit is less than 1, the series converges absolutely. If the limit is greater than 1 or the limit does not exist, the series diverges. If the limit is equal to 1, the test is inconclusive.
- Determine the interval of convergence by considering the endpoints separately, testing convergence at each endpoint.

The series converges for ( -\frac{1}{2} < x \leq \frac{1}{2} ). At ( x = -\frac{1}{2} ) and ( x = \frac{1}{2} ), further analysis is required to determine convergence or divergence.

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When evaluating a one-sided 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 one-sided 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 one-sided 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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