# What is the radius of convergence of #sum_(n=1)^oo sin(x/n^2)#?

is absolutely convergent with radius of convergence

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To find the radius of convergence of the series (\sum_{n=1}^\infty \sin\left(\frac{x}{n^2}\right)), we can use the ratio test. The radius of convergence, denoted by (R), is the limit of the absolute value of the ratio of consecutive terms as (n) approaches infinity. If this limit exists, it equals (R).

Applying the ratio test to the given series:

[a_n = \sin\left(\frac{x}{n^2}\right)]

[a_{n+1} = \sin\left(\frac{x}{(n+1)^2}\right)]

[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{\sin\left(\frac{x}{(n+1)^2}\right)}{\sin\left(\frac{x}{n^2}\right)}\right|]

Now, as (n) approaches infinity, both (\frac{1}{n^2}) and (\frac{1}{(n+1)^2}) approach zero. Therefore, (\sin\left(\frac{x}{n^2}\right)) and (\sin\left(\frac{x}{(n+1)^2}\right)) approach (x) and (\sin(x)), respectively.

[\lim_{n \to \infty} \left|\frac{\sin(x)}{\sin(x)}\right| = 1]

Since the limit is 1, the ratio test is inconclusive. However, the series (\sum_{n=1}^\infty \sin\left(\frac{x}{n^2}\right)) converges for all (x) by the Dirichlet's Test. Hence, the radius of convergence is infinite.

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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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