What is the least upper bound of the sequence defined by the following piecewise function:
$begingroup$
What is the least upper bound of the sequence defined, for $ninmathbb{N}$, by
$$a_{n}=begin{cases}
3/n, & text{if }ntext{ is odd;}\
1/n, & text{if }ntext{is even.}
end{cases}$$
I know that the least upper bound is $3$, but how do I determine the least upper bound?
calculus upper-lower-bounds
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
$endgroup$
add a comment |
$begingroup$
What is the least upper bound of the sequence defined, for $ninmathbb{N}$, by
$$a_{n}=begin{cases}
3/n, & text{if }ntext{ is odd;}\
1/n, & text{if }ntext{is even.}
end{cases}$$
I know that the least upper bound is $3$, but how do I determine the least upper bound?
calculus upper-lower-bounds
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
$endgroup$
add a comment |
$begingroup$
What is the least upper bound of the sequence defined, for $ninmathbb{N}$, by
$$a_{n}=begin{cases}
3/n, & text{if }ntext{ is odd;}\
1/n, & text{if }ntext{is even.}
end{cases}$$
I know that the least upper bound is $3$, but how do I determine the least upper bound?
calculus upper-lower-bounds
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
$endgroup$
What is the least upper bound of the sequence defined, for $ninmathbb{N}$, by
$$a_{n}=begin{cases}
3/n, & text{if }ntext{ is odd;}\
1/n, & text{if }ntext{is even.}
end{cases}$$
I know that the least upper bound is $3$, but how do I determine the least upper bound?
calculus upper-lower-bounds
calculus upper-lower-bounds
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
edited Dec 7 '18 at 3:21
zipirovich
11.2k11631
11.2k11631
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
asked Dec 7 '18 at 2:33
hunnybunshunnybuns
15
15
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Notice the behavior of the sequence as $n to infty$ on even and odd $n$ separately. In each case, for every even $n$, the sequence is monotonic decreasing, and the same for every odd $n$ the same is also true.
Logically, then, this suggests that the sequences' highest points are at the start of the sequence, i.e. $n=1$ for odd $n$ and $n=2$ for even $n$. Well, in those cases, you get the values $a_1 = 3$ and $a_2 = 1/2$.
The former is visibly higher, so we say that the least upper bound is $3$ - it is the least element such that each subsequent element is less than it.
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Notice the behavior of the sequence as $n to infty$ on even and odd $n$ separately. In each case, for every even $n$, the sequence is monotonic decreasing, and the same for every odd $n$ the same is also true.
Logically, then, this suggests that the sequences' highest points are at the start of the sequence, i.e. $n=1$ for odd $n$ and $n=2$ for even $n$. Well, in those cases, you get the values $a_1 = 3$ and $a_2 = 1/2$.
The former is visibly higher, so we say that the least upper bound is $3$ - it is the least element such that each subsequent element is less than it.
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
add a comment |
$begingroup$
Notice the behavior of the sequence as $n to infty$ on even and odd $n$ separately. In each case, for every even $n$, the sequence is monotonic decreasing, and the same for every odd $n$ the same is also true.
Logically, then, this suggests that the sequences' highest points are at the start of the sequence, i.e. $n=1$ for odd $n$ and $n=2$ for even $n$. Well, in those cases, you get the values $a_1 = 3$ and $a_2 = 1/2$.
The former is visibly higher, so we say that the least upper bound is $3$ - it is the least element such that each subsequent element is less than it.
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
add a comment |
$begingroup$
Notice the behavior of the sequence as $n to infty$ on even and odd $n$ separately. In each case, for every even $n$, the sequence is monotonic decreasing, and the same for every odd $n$ the same is also true.
Logically, then, this suggests that the sequences' highest points are at the start of the sequence, i.e. $n=1$ for odd $n$ and $n=2$ for even $n$. Well, in those cases, you get the values $a_1 = 3$ and $a_2 = 1/2$.
The former is visibly higher, so we say that the least upper bound is $3$ - it is the least element such that each subsequent element is less than it.
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
$endgroup$
Notice the behavior of the sequence as $n to infty$ on even and odd $n$ separately. In each case, for every even $n$, the sequence is monotonic decreasing, and the same for every odd $n$ the same is also true.
Logically, then, this suggests that the sequences' highest points are at the start of the sequence, i.e. $n=1$ for odd $n$ and $n=2$ for even $n$. Well, in those cases, you get the values $a_1 = 3$ and $a_2 = 1/2$.
The former is visibly higher, so we say that the least upper bound is $3$ - it is the least element such that each subsequent element is less than it.
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
answered Dec 7 '18 at 2:39
Eevee TrainerEevee Trainer
5,7961936
5,7961936
add a comment |
add a comment |
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