The Integral test is applicable if f is positive and decreasing function on the infinite interval `[k, oo)` where` kgt= 1` and `a_n=f(x)` . Then the series `sum_(n=1)^oo a_n` converges if and only if the improper integral `int_1^oo f(x) dx` converges. If the integral diverges then the series also diverges.

For the given series `sum_(n=1)^oo 1/n^(1/2)` , then `a_n = 1/n^(1/2)` then applying `a_n=f(x)` , we consider:

`f(x) = 1/x^(1/2)` .

As shown on the graph of `f(x)` , the function is positive on the interval `[1,oo)` . As x at the denominator side gets larger, the function value decreases.

Therefore, we may determine the convergence of the improper integral as:

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

Apply Law of exponent: `1/x^m = x^(-m)` .

`lim_(t-gtoo)int_1^t 1/x^(1/2) dx =lim_(t-gtoo)int_1^t x^(-1/2) dx`

Apply Power rule for integration:` int x^n dx = x^(n+1)/(n+1)` .

`lim_(t-gtoo)int_1^t x^(-1/2) dx=lim_(t-gtoo)[ x^(-1/2+1)/(-1/2+1)]|_1^t`

`=lim_(t-gtoo)[ x^(1/2)/(1/2)]|_1^t`

`=lim_(t-gtoo)[ x^(1/2)*(2/1)]|_1^t`

`=lim_(t-gtoo)[ 2x^(1/2)]|_1^t`

or `lim_(t-gtoo)[ 2sqrt(x)]|_1^t`

Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`lim_(t-gtoo)[ 2sqrt(x)]|_1^t=lim_(t-gtoo)[2sqrt(t) -2sqrt(1)]`

`=lim_(t-gtoo)[2sqrt(t) -2*1]`

`=lim_(t-gtoo)[2sqrt(t) -2]`

`= oo`

Note: `lim_(t-gtoo)( -2) =-2` and `lim_(t-gtoo)2sqrt(t) = oo ` then `oo-2~~oo` .

The` lim_(t-gtoo)[ 2sqrt(x)]|_1^t = oo` implies that the integral diverges.

**Conclusion**: The integral`int_1^oo 1/x^(1/2) ` diverges,
therefore the series `sum_(n=1)^oo 1/n^(1/2) ` must also diverge.

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