Links

Basic Examples

Proofs for the problems below follow a common structure, as described on the prior page.
We strongly encourage readers attempt each problem by first drawing a picture, and then attempting a formal write-up before reading the example write-up provided below.
Example 1.
Define the sequence
{an}\{a_n\}
by
annn2+9a_n \triangleq \dfrac{n}{n^2+9}
for all
nN.n\in\mathbb{N}.
Prove
{an}\{a_n\}
converges to zero.
Ex 1: Comment
Hint
Visual
Formal Write-Up
Follow-Up
Although the result may be clear from calculus, the point here is to provide formal arguments.
Note
0<an<1n.0 < a_n < \dfrac{1}{n}.
Plot of and . We see , and so as . Also note for large .
We claim
{an}\{a_n\}
converges to zero and prove this as follows. Let
ε>0\varepsilon > 0
be given. It suffices to show there exists
NNN\in\mathbb{N}
such that
an0ε|a_n - 0 | \leq\varepsilon
for all
nN.n\geq N.
For each
nNn\in \mathbb{N}
, note
n>0n > 0
and so
1/n>01/n > 0
, which implies
n<n+9n        1<n+9nn        1n+9n<1n.n < n + \dfrac{9}{n} \ \ \implies \ \ 1 < \dfrac{n + \frac{9}{n}}{n} \ \ \implies \ \ \dfrac{1}{n + \frac{9}{n}} < \dfrac{1}{n}.
Consequently,
where the second equality holds since
n>0n > 0
and
n2+9>0n^2 + 9 > 0
. By the Archimedean Property of
R\mathbb{R}
, there exists
N~N\tilde{N}\in\mathbb{N}
such that
1<N~ε1 < \frac{\tilde{N}}{\varepsilon}
, which implies
1N~<ε\frac{1}{\tilde{N}}< \varepsilon
. Thus, combining the above results,
which verifies the desired inequality, taking
N=N~.N=\tilde{N}.
\blacksquare
  • We emphasize each statement is a complete sentence (creativity is not required).
  • It is completely acceptable to mirror this structure, but change the inequalities/terms for another problem.
  • Readability is greatly aided by concluding with a statement that explicitly states how an initial task (which we set out to do) was completed.
Example 2.
Define the sequence
{an}\{a_n\}
​ by
4n3+nn3+6\dfrac{4n^3+n}{n^3+6}
​ for all
nN.n\in\mathbb{N}.
​ Prove
{an}\{a_n\}
converges.​
Ex 2: Comment
Hint
Visual
Formal Write-Up
We can show
{an}\{a_n\}
​ converges by verifying it converges to a limit (that you should find).
As
nn
​ gets big, the dominant term in the numerator is
4n34n^3
​ and in the denominator is
n3n^3
​, which should give insight about the limit of
{an}.\{a_n\}.
Animation of the sequence converging.
We claim
{an}\{a_n\}
​ converges to four and prove this as follows. Let
ε>0\varepsilon>0
​ be given. If suffices to show there is
NNN\in\mathbb{N}
​ such that
an0ε|a_n-0|\leq\varepsilon
​ for all
nN.n\geq N.
​ For each
nNn\in\mathbb{N}
​, observe
an4=4n3+nn3+64=(4n3+n)(n3+6)4n3+6=n24n3+6.|a_n-4|=\left|\dfrac{4n^3+n}{n^3+6}-4\right|=\left| \dfrac{(4n^3+n)-(n^3+6)4}{n^3+6}\right|=\left|\dfrac{n-24}{n^3+6}\right|.
Building on this inequality,
n24n\geq 24
​ implies
an4=n24n3+6nn3+6nn3nn2=1n,|a_n - 4| = \dfrac{n-24}{n^3+6} \leq \dfrac{n}{n^3+6} \leq \dfrac{n}{n^3} \leq \dfrac{n}{n^2}=\dfrac{1}{n},
​where we note
n2=nnn1=nn^2=n\cdot n \geq n\cdot 1=n
​ implies
1/n21/n.1/n^2\leq 1/n.
​ By the Archimedean property of
R\mathbb{R}
​, there exists a natural number
N1NN_1\in\mathbb{N}
​ such that
1N1ε1 \leq N_1 \varepsilon
​, and so
1/N1ε1/N_1 \leq \varepsilon
​. Setting
N2=max(24, N1)N_2 = \max(24,\ N_1)
yields, by the above results,
This verifies the desired inequality, taking
N=N2.N=N_2.
\blacksquare
Example 3.
Suppose
{cn}\{c_n\}
​ is a sequence in
[3,7)[-3,7)
, i.e.
1cn<7-1\leq c_n<7
for all
nN.n\in\mathbb{N}.
​ Define the sequence
{an}\{a_n\}
​ by
ancn/na_n\triangleq c_n/n
​ for all
nN.n\in\mathbb{N}.
​ Prove
limnan=0.\lim_{n\rightarrow\infty} a_n = 0.
Ex 3: Comment
Hint
Visual
Formal Write-Up
Follow-Up
Be careful to formally state what must be shown (to ensure no confusion between the use of each of the listed sequences).
Here is it helpful to bound
ana_n
​ in terms of a bound for
cnc_n
​.
Example sequence convergence within bounds for 7/n and -7/n.
Let
ε>0\varepsilon > 0
​ be given. It suffices to show there is
NNN\in\mathbb{N}
​ such that
an0ε|a_n - 0|\leq \varepsilon
​ for all
nN.n\geq N.
​ By our hypothesis on
{cn}\{c_n\}
​,
By the Archimedean property of
R\mathbb{R}
​, there is
N1NN_1\in\mathbb{N}
​ such that
7<N1ε7 < N_1 \varepsilon
​, which implies
7/N1<ε.7/N_1 < \varepsilon.
​ Hence
This shows the desired result, taking
N=N1N=N_1
​.
\blacksquare
A trend in the previous three problems is to use the Archimedean property of
R\mathbb{R}
​. This is not a coincidence! Be sure to know how to use this property so you may obtain the existence of the natural number needed in the definition of limit convergence.

Exercises.

Exercise 1. Prove the sequence
{an}\{a_n\}
defined by
an2n2+73+n2a_n\triangleq \dfrac{2n^2+7}{3+n^2}
converges to two.
  • State the definition of what must be shown.
  • Rewrite the inequality with
    an2.a_n - 2.
  • Apply the Archimedean property of
    R\mathbb{R}
    .
  • Combine relations to verify the inequality that must be shown.​