Since 1∕n → 0 then 17∕n → 0 and 2∕n → 0. Therefore 5 + 17∕n → 5 and 1 + 2∕n → 1. Thus → 1 implying that
→ 5. Given the opening equation, this then final implies
Since 1∕n → 0, then 1∕n2 → 0. This implies 3 - 2∕n + 17∕n2 → 3 and 1 + 21∕n + 2∕n2 → 1. Hence converges
to three and thus so does
, as per the opening equation.
Hence, given the opening equation of this problem, we have - n →
Thus limn→∞an ≤ 2mam limn→∞ = 0
but for any integer N, ≤
+
so that
where as
![]() | (6.1) |
Assume that both tn → s and rn → s. Then for any r > 0, there is are integers N1,N2 such that rn,tn ∈ B(s) for all
n ≥ N where N = max(N1,N2). Therefore |rn - tn| <
which, combined with equation 6.1, implies |sn - tn| <
for
each n ≥ N. Therefore
for each n ≥ N. Hence Br(s) contains all but finitely many points in {sn}, so that sn → s.
be the fraction of natural numbers from one to n which are not divisible by three. Thus defining the sequence {tn} where tn is the fraction of natural numbers less than or equal to n that are divisibile by three, we have
![]() | (7.2) |
for each n. Now tn = , which implies tn ≤
for each n. Thus for any ε > 0 we have
so that tn →. Thus due to Equation 7.2, sn →
.
Repeat this for all natural numbers not multiples of two or three
Let {tn} again be as above, the fraction of natural numbers less than or equal to n which are divisible by three. Define
{rn} as the sequence where each rn is the fraction of natural numbers less than or equal to n which are divisible by two.
Thus we have rn = , which with an argument nearly identical to the one for {tn} converging to
above, we know
rn →
. If, now, the sequence {an} is such that each an is the fraction of natural numbers less than or equal to n which are
not divisible by two or three, we have
Hence an converges to =
since tn and rn converge to
and
, respectively.
We note that for the real-valued function f(x) = we have f′(x) =
. Thus f′(x) = 0 when x =
,
implying that f(
) =
is the minimum value of f. Since {xn} are just particular values of f, then {xn} is bounded
below by
.
So let’s define a sequence {yn} by yn = xn -. Then we have
This then indicates that yn+1 < <
<
which in turn implies that yn → 0. Since we defined yn = xn -
, then
xn = yn +
yielding xn →
.
diverges. This implies that
![]() | (9.3) |
also diverges since it’s one less than the first. We use this fact to our advantage in finding a counter-example. Define the sequence {an} by recursively by a0 = 0 and
This sequence essentially bounces back and forth between zero and one, so is bounded. It also satisfies the property that
|an -an-1| < since each point an differs by
from the previous. Furthermore, we know once the sequence “hits” one
and heads back to zero (and vice versa), that it will actually reach zero because of the divergence of 9.3 above; i.e. starting
at zero but not being able to reach one (and vice versa) at any point in the sequence would imply convergence of
9.3.
converging to zero.
![]() | (10.4) |
Furthermore, because {an} converges, it is bounded, so that we can find an integer M with
![]() | (10.5) |
for all n. Thus for n > N we have
implying that
implying the convergence of {cn} to A.
which implies cn → 0.