# Determine whether the following series converges or diverges `sum_(n=1)^oo 1/sqrt n`

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### 1 Answer

We may use integral test to determine if the series is convergent or divergent.

`sum_(n=1)^oo 1 /sqrtn`

In integral form, we may represent this as:

`int_1^oo 1/sqrtx dx`

To evaluate the integral, replace the upper limit with t variable. Then, take the limit of the integral as t approaches infinity. So we have,

`= lim_(t->oo) int_1^t 1/sqrtx dx = lim_(t->oo) int_1^t x^(-1/2) dx`

`= lim_(t->oo) 2x^(1/2)` `|_1^t = lim_(t->oo) (2t^(1/2) -2) = lim_(t->oo)2t^(1/2) - lim_(t->oo) 2`

Let's try to take the limit of each separately.

To evaluate, use rule `lim_(x->oo) x^c=oo` . Also, take note that if we multiply a large number with a small one, the product is a large number.

`lim_(t->oo) 2t^(1/2) =2*oo = oo`

For the second one, note that a limit of a constant is equal to the constant itself.

`lim_(x->oo) 2 = 2`

So we have,

= `lim_(t->oo) 2t^(1/2) - lim_(t->oo)2 = oo - 2=oo`

The limit is infinite since subtracting a small number from a large value result to a large value.

So the result of the integral test is:

`int_1^oo 1/sqrtxdx=oo`

*This indicates that the integral is divergent.*

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**Since the integral diverges, therefore the series `sum_(n=1)^oo 1/sqrtn ` diverges too.**