Download Math 256 Quiz 6 Solutions

Math 256
Quiz 6 Solutions
Calculus II
Kate Bella
1) Find a formula for the nth term of the sequence
1 2 3 4
0, , , , ..... .
4 6 8 10
We know that we need each term of the sequence to be calculated with a formula using its position as an
input. Notice that the numerator is one less than the position of the term. Also, we see that the denominator
is always an even number. We want this set of even numbers to start at 2, which means that it will always
be double the position.
So we have the formula an =
n−1
.
2n
2) Find the limit of the sequence
2n + 1
.
3n
We can consider the limit of the corresponding rational function, and apply l’Hopital’s rule.
lim
n→∞
3) Show that the sequence
3
n2
2n + 1
2
2
= lim
=
n→∞ 3
3n
3
converges by showing that it is bounded and monotonic.
First, we can see that this function is decreasing with the following inequalities.
n
<
n+1
2
<
(n + 1)2
1
(n + 1)2
3
(n + 1)2
n
1
n2
3
n2
>
>
Also, notice that all of the terms will be positive since the denominator is squared, so there is a lower
boundary of 0. The upper bound will be 3 since this is the first value of the sequence, and we already know
it is strictly decreasing. Thus the sequence is bounded.