September 29, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework04

We first see that

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

Similar to the previous part, we first see

Since 1∕n → 0, then 1∕n^{2} → 0. This implies 3 - 2∕n + 17∕n^{2} → 3 and 1 + 21∕n + 2∕n^{2} → 1. Hence converges
to three and thus so does , as per the opening equation.

We first note that

Hence, given the opening equation of this problem, we have - n →

Let {a

Fix an integer m so that m > 2c. The for any n > m we have that

Thus lim_{n→∞}a_{n} ≤ 2^{m}a_{m} lim_{n→∞} = 0

Let {a

By definition we have

but for any integer N, ≤ + so that

Let {a

where as

Let {r

| (6.1) |

Assume that both t_{n} → s and r_{n} → s. Then for any r > 0, there is are integers N_{1},N_{2} such that r_{n},t_{n} ∈ B_{}(s) for all
n ≥ N where N = max(N_{1},N_{2}). Therefore |r_{n} - t_{n}| < which, combined with equation 6.1, implies |s_{n} - t_{n}| < for
each n ≥ N. Therefore

for each n ≥ N. Hence B_{r}(s) contains all but finitely many points in {s_{n}}, so that s_{n} → s.

Let A be the set of natural numbers not divisible by three. Define

be the fraction of natural numbers from one to n which are not divisible by three. Thus defining the sequence {t_{n}} where
t_{n} 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 t_{n} = , which implies t_{n} ≤ for each n. Thus for any ε > 0 we have

so that t_{n} →. Thus due to Equation 7.2, s_{n} →.

Repeat this for all natural numbers not multiples of two or three
Let {t_{n}} again be as above, the fraction of natural numbers less than or equal to n which are divisible by three. Define
{r_{n}} as the sequence where each r_{n} is the fraction of natural numbers less than or equal to n which are divisible by two.
Thus we have r_{n} = , which with an argument nearly identical to the one for {t_{n}} converging to above, we know
r_{n} →. If, now, the sequence {a_{n}} is such that each a_{n} is the fraction of natural numbers less than or equal to n which are
not divisible by two or three, we have

Hence a_{n} converges to = since t_{n} and r_{n} converge to and , respectively.

Define {x

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 {x_{n}} are just particular values of f, then {x_{n}} is bounded
below by .

So let’s define a sequence {y_{n}} by y_{n} = x_{n} -. Then we have

This then indicates that y_{n+1} < < < which in turn implies that y_{n} → 0. Since we defined y_{n} = x_{n} -, then
x_{n} = y_{n} + yielding x_{n} →.

This is not true. We know that the series

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 {a_{n}} by recursively by a_{0} = 0 and

This sequence essentially bounces back and forth between zero and one, so is bounded. It also satisfies the property that
|a_{n} -a_{n-1}| < since each point a_{n} 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.

For a real sequence {a

Let a

converging to zero.

Let a

| (10.4) |

Furthermore, because {a_{n}} 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 {c_{n}} to A.

No. For example, let {a

which implies c_{n} → 0.