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