# How do you determine the convergence or divergence of #Sigma 1/nsin(((2n-1)pi)/2)# from #[1,oo)#?

It is convergent.

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To determine the convergence or divergence of the series ( \sum_{n=1}^{\infty} \frac{1}{n\sin\left(\frac{(2n-1)\pi}{2}\right)} ) from ( n = 1 ) to infinity, follow these steps:

- Examine the behavior of the terms of the series as ( n ) approaches infinity.
- Note that ( \sin\left(\frac{(2n-1)\pi}{2}\right) ) oscillates between -1 and 1 as ( n ) increases, and it approaches 0 when ( n ) is odd.
- For odd values of ( n ), ( \sin\left(\frac{(2n-1)\pi}{2}\right) = 0 ), which makes the corresponding term of the series undefined.
- Consequently, when ( n ) is odd, the terms of the series approach infinity, indicating divergence.
- However, when ( n ) is even, ( \sin\left(\frac{(2n-1)\pi}{2}\right) = \pm 1 ), leading to bounded terms.
- Since the series contains infinitely many terms for both even and odd ( n ), and the terms are unbounded for odd ( n ), the series diverges.

Therefore, the given series diverges.

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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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