Determine whether the series converge or diverge?

  1. #\sum_(n=1)^\inftyn#
    is divergent; why?
  2. #\sum_(n=1)^\infty(-1)^n#
    is divergent; why?
  3. #\sum_(n=1)^\infty(1)/(3^(n-1))#
    is convergent #S=3/2#; why?
  4. #\sum_(n=1)^\infty(1)/(3^n)#
    is convergent #S=2/9#; why?
  5. #\sum_(n=1)^\infty(4n^2-n^3)/(10+2n^3)#
    is divergent; why?
  6. #\sum_(n=1)^\infty(2/3^n+2/(3n))#
    is divergent; why?
  7. #\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!)

1 Answer
Apr 28, 2017

(2) is not precise and (4) is incorrect

Explanation:

(1) The series does not respect Cauchy's necessary condition since:

#lim_(n->oo) n = oo#

then it cannot converge.

In fact the partial sum #s_N# is:

#s_N = sum_(n=1)^N n = 1+2+3+...+N = (N(N-1))/2#

so:

#lim_(N->oo) s_N = oo#

(2) Again, the series does not respect Cauchy's necessary condition since:

#lim_(n->oo) (-1)^n#

does not exist.

In fact if #s_N# is the partial sum we can see that:

#s_1 = -1#

#s_2 = 0#

and we can prove by induction that:

#s_N = {(-1 " for " N " odd "),(0 " for " N " even "):}#

so that #lim_(N->oo) s_N# does not exist ( and then the series is undetermined, not divergent).

(3) This can be reduced to the sum of a geometric series, and we know that:

#sum_(n=0)^oo a^n = 1/(1-a)# for #abs a < 1#

In fact we have:

#sum_(n=1)^oo 1/3^(n-1)#

substitute #k= n-1#

#sum_(n=1)^oo 1/3^(n-1) = sum_(k=0)^oo 1/3^k = sum_(k=0)^oo (1/3)^k = 1/(1-1/3) = 3/2#

(4) As above:

#sum_(n=1)^oo 1/3^n#

is just a geometric series lacking the first term: if we add and subtract #1# and note that #(1/3)^0 = 1# we get:

#sum_(n=1)^oo 1/3^n = -1 + 1 + sum_(n=1)^oo 1/3^n = -1 + sum_(n=0)^oo (1/3)^n = -1+1/(1-1/3) = -1+3/2 = 1/2#

You can also solve it noting that:

#sum_(n=1)^oo 1/3^n = 1/3 sum_(n=1)^oo 1/3^(n-1)#

so from the previous exercise:

#sum_(n=1)^oo 1/3^n = 1/3 sum_(n=1)^oo 1/3^(n-1) = 1/3 xx 3/2 = 1/2#

(5) We have:

#sum_(n=1)^oo (4n^2-n^3)/(10+2n^3)#

Also this series does not satisfy Cauchy's necessary condition as:

#lim_(n->oo) (4n^2-n^3)/(10+2n^3) = -1/2#

In fact if we divide numerator and denominator by #n^3#:

#a_n = (4/n-1)/(10/n^3+2 ) #

For #n > 4# we have that #a_n# is negative so if we decrease the denominator we have a negative quantity that is larger in absolute value, that is:

#a_n = (4/n-1)/(10/n^3+2 ) < -1/2 (1- 4/n)# for #n > 4#

Now for #n > 8# we have:

#4/n < 4/8#

so:

#1-4/n > 1-1/2#

#1-4/n > 1/2# for #n > 8#

and then:

#a_n < -1/4# for #n > 8#

Now consider the partial sum for #N < 8#:

#s_N = s_7 + sum_(n=8)^N a_n < s_7 -(N-8)/4#

and since #lim_(N->oo) s_7 -(N-8)/4 = -oo#

then also:

#lim_(N->oo) s_N = -oo#

(6) We have:

#sum_(n=1)^oo (2/3^n+2/(3n)) = 2/3 sum_(n=1)^oo (1/3^(n-1)+1/n)#

but then:

#a_n = 1/3^(n-1)+1/n > 1/n#

and as the harmonic series is divergent then by comparison also this series is divergent.

(7) Again a geometric series:

#sum_(n=1)^oo 9^(2−n)4^(n+1) = 4*4*9 sum_(n=1)^oo 4^(n-1)/9^(n-1) = 144 sum_(n=1)^oo (4/9)^(n-1) = 144*1/(1-4/9) =9/5*144 = 259.2#