`sum_(n=1)^oo 1/n^3` Use the Integral Test to determine the convergence or divergence of the p-series.

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marizi eNotes educator| Certified Educator

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.