Determine whether the series converge or diverge?
\sum_(n=1)^\inftyn
is divergent; why?
\sum_(n=1)^\infty(-1)^n
is divergent; why?
\sum_(n=1)^\infty(1)/(3^(n-1))
is convergent S=3/2 ; why?
\sum_(n=1)^\infty(1)/(3^n)
is convergent S=2/9 ; why?
\sum_(n=1)^\infty(4n^2-n^3)/(10+2n^3)
is divergent; why?
\sum_(n=1)^\infty(2/3^n+2/(3n))
is divergent; why?
\sum_(n=1)^\infty9^(2-n)4^(n+1)
is convergent S=259.2 ; why?
I would mainly like to know the best method of testing convergence/divergence in each of these problems. I don't want to waste time on a work-intensive process during a test...
(Please don't teach shortcuts unless they are proven to work!)
\sum_(n=1)^\inftyn
is divergent; why?\sum_(n=1)^\infty(-1)^n
is divergent; why?\sum_(n=1)^\infty(1)/(3^(n-1))
is convergentS=3/2 ; why?\sum_(n=1)^\infty(1)/(3^n)
is convergentS=2/9 ; why?\sum_(n=1)^\infty(4n^2-n^3)/(10+2n^3)
is divergent; why?\sum_(n=1)^\infty(2/3^n+2/(3n))
is divergent; why?\sum_(n=1)^\infty9^(2-n)4^(n+1)
is convergentS=259.2 ; why?
I would mainly like to know the best method of testing convergence/divergence in each of these problems. I don't want to waste time on a work-intensive process during a test...
(Please don't teach shortcuts unless they are proven to work!)
1 Answer
(2) is not precise and (4) is incorrect
Explanation:
(1) The series does not respect Cauchy's necessary condition since:
then it cannot converge.
In fact the partial sum
so:
(2) Again, the series does not respect Cauchy's necessary condition since:
does not exist.
In fact if
and we can prove by induction that:
so that
(3) This can be reduced to the sum of a geometric series, and we know that:
In fact we have:
substitute
(4) As above:
is just a geometric series lacking the first term: if we add and subtract
You can also solve it noting that:
so from the previous exercise:
(5) We have:
Also this series does not satisfy Cauchy's necessary condition as:
In fact if we divide numerator and denominator by
For
Now for
so:
and then:
Now consider the partial sum for
and since
then also:
(6) We have:
but then:
and as the harmonic series is divergent then by comparison also this series is divergent.
(7) Again a geometric series: