`sum_(n=1)^oo 1/n^5` Use the Integral Test to determine the convergence or divergence of the p-series.

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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.

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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^5` , the `a_n = 1/n^5 ` then applying `a_n=f(x),` we consider:

`f(x) = 1/x^5` .  

As shown on the graph for 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^5 = lim_(t-gtoo)int_1^t 1/x^5 dx`

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

`lim_(t-gtoo)int_1^t 1/x^5 dx =lim_(t-gtoo)int_1^t x^(-5) dx`

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

`lim_(t-gtoo)int_1^t 1/x^5 dx =lim_(t-gtoo)[ x^(-5+1)/(-5+1)]|_1^t`

                            ` =lim_(t-gtoo)[ x^(-4)/(-4)]|_1^t`

                            ` =lim_(t-gtoo)[ -1/(4x^4)]|_1^t`

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

`lim_(t-gtoo)[ -1/(4x^4)]|_1^t=lim_(t-gtoo)[-1/(4*t^4) -(-1/(4*1^4))]`

                             `=lim_(t-gtoo)[(-1/(4t^4))-(-1/4 )]`

                            `=lim_(t-gtoo)[-1/(4t^4)+1/4]`

                           `= 1/4 or0.25`

Note: `lim_(t-gtoo)[1/4] =1/4 ` and `lim_(t-gtoo)1/(4t^4) = 1/oo or 0`

The integral `int_1^oo 1/x^5`  is convergent therefore the p-series `sum_(n=1)^oo 1/n^5 ` must also be convergent. 

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