# How to do comparison test for #sum 1 / (sqrt(n))# n=1 to #n=oo#?

The series diverges.

Let's begin with an inequality :

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To apply the comparison test for the series ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} ), we need to find a series whose terms are always less than or equal to the terms of the given series, and whose convergence or divergence we know.

Given that ( \frac{1}{\sqrt{n}} ) is positive for all ( n ), we can compare it to the series ( \sum_{n=1}^{\infty} \frac{1}{n^{p}} ), where ( p > 0 ). Specifically, we can choose ( p = \frac{1}{2} ) since ( \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} ).

We know that the series ( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} ) is a p-series with ( p = \frac{1}{2} ), and it converges because ( p > 0 ).

Therefore, by the comparison test, since ( \frac{1}{\sqrt{n}} ) is always less than or equal to ( \frac{1}{n^{1/2}} ) for all ( n ), and ( \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} ) converges, then ( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} ) also converges.

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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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