Respuesta :
Answer:
Hence series can be determine as convergent since the function is positive and decreasing on [1,infinity].
Step-by-step explanation:
Given: the function f(x)=5cos(πx)/x as series from [1,infinity]
To find : Is series positive or decreasing and converges in given range [1,infinity].
Solution:
we have series as : [1,infinity] with function 5cos(πn)/n and 5 being constant
consider the dependent function cos(πn) and 1/n we get ,
by definition cos(πn)=(-1)^n .
hence ,
summation as n[1,infinity] function as [tex]\frac{(-1)^n}{n}[/tex].
using alternate series test, series converges as 1/n tends to 0 and decreases ,but
by integral test is not convergent series because :
Sn=1+1/2+1/3+1/4+.........+1/n > integral with limits (1 to n+1) with function (1/x)dx=ln(x) with (1 to n+1) .
hence =ln(n+1)
as n tends to infinity n+1 will be tending infinity.
It is harmonic series ,
[tex]\lim_{n \to \infty} \int\limits^({1/x} \, dx[/tex] =infinity ,with limits as (1 to n+1)
[tex]\int\limits^a_b {(1/x)} \, dx[/tex] =infinity. with limits as (1 to n+1).
hence we can prove that series convergent or divergent with improper integral .It is called as integral test .
Hence we cannot use the integral test.
Integral Test:
In the mathematical domain, the Integral test for convergence is a technique that is often applied for the purpose of testing an infinite series of non-negative terms for convergence.
Given function is,
[tex]f(x) = 5 cos(\pi x)[/tex]
This is an alternating series.
So we need to do the alternating series test, not the integral test.
The integral test is for the positive series, while this series alternate between positive and negative.
Hence we cannot use the integral test.
Learn more about the topic Integral Test:
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