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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