Recall that 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^3` , the `a_n = 1/n^3` then applying `a_n=f(x)` , we consider:
`f(x) = 1/x^3` . The function...
Recall that 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^3` , the `a_n = 1/n^3` then applying `a_n=f(x)` , we consider:
`f(x) = 1/x^3` . The function is positive and 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^3 = lim_(t-gtoo)int_1^t 1/x^3 dx`
Apply the Law of exponent: `1/x^m = x^(-m)` .
`lim_(t-gtoo)int_1^t 1/x^3 dx =lim_(t-gtoo)int_1^t x^(-3) dx`
Apply Power rule for integration: `int x^n dx = x^(n+1)/(n+1).`
`lim_(t-gtoo)int_1^t 1/x^3 dx =lim_(t-gtoo)[ x^(-3+1)/(-3+1)]|_1^t`
` =lim_(t-gtoo)[ x^(-2)/(-2)]|_1^t`
`=lim_(t-gtoo)[ -1/(2x^2)]|_1^t`
Apply the definite integral formula: `F(x)|_a^b = F(b)-F(a)` .
`lim_(t-gtoo)[ -1/(2x^2)]|_1^t=lim_(t-gtoo)[-1/(2*t^2) -(-1/(2*1^2))]`
`=lim_(t-gtoo)[ -1/(2t^2)-(-1/2)]`
`=lim_(t-gtoo)[-1/(2t^2)+1/2]`
`= 1/2` .
Note:` lim_(t-gtoo) 1/2 =1/2 ` and `lim_(t-gtoo)1/(2t^2) = 1/oo or 0`
The integral `int_1^oo 1/x^3` is convergent therefore the series `sum_(n=1)^oo 1/n^3` must also be convergent.
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