Question

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!)

Answer 1

(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`