# How do you find the interval of convergence #Sigma x^n/n^2# from #n=[1,oo)#?

The interval of convergence is

Use the Ratio test to compute the convergence interval

Therefore,

The series converges absolutely

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To find the interval of convergence for the series Σ(x^n/n^2) from n=1 to infinity, you can use the ratio test.

First, apply the ratio test:

- Take the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term.

The series converges if the limit is less than 1 and diverges if the limit is greater than 1.

In this case, the series is Σ(x^n/n^2). So, applying the ratio test:

- Take the absolute value of the ratio of the (n+1)th term to the nth term, which gives ((x^(n+1))/(n+1)^2) / ((x^n)/n^2).
- Simplify the expression to get ((x^(n+1))(n^2))/((x^n)((n+1)^2)).
- Cancel out the common terms to get x/(1 + 2/n), which simplifies to x as n approaches infinity.

Since the limit of x as n approaches infinity is x, the series will converge for |x| < 1.

Therefore, the interval of convergence is -1 < x < 1.

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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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