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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