Monotonicity of $a_n=1+frac{(-1)^n}{n}$
$begingroup$
I'm trying to study the monotonicity of $a_n=1+frac{(-1)^n}{n}$, but what I'm getting isn't correct:
I just assume that $a_n$ is monotonically increasing, and if it isn't, I'll get something absurd:
So, $a_n$ is monotonically increasing iff
$$
begin{split}
a_{n+1} geq a_n
&iff left(1+frac{(-1)^{n+1}}{n+1}right)
- left(1+frac{(-1)^{n}}{n}right) geq 0 \
&iff frac{(-1)^{n+1}}{n+1} - frac{(-1)^{n}}{n} geq 0 \
&iff frac{(-1)^{n+1}}{n+1} geq frac{(-1)^{n}}{n},
end{split}
$$
Which doesn't make sense since $lim_{n to infty} frac{(-1)^n}{n}=0$ and $lim_{n to infty} frac{(-1)^{n+1}}{n+1}=0$, but the later one decreases more rapidly, so $a_n$ would be monotonically decreasing.
But, the solution says that if $n$ is even then $a_n$ is monotonically decreasing and if $n$ is odd then monotonically increasing. How do I prove that? What's wrong with what I did?
real-analysis calculus sequences-and-series monotone-functions
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
asked Dec 24 '18 at 15:51
E----E----
32410
$endgroup$
add a comment |
$begingroup$
I'm trying to study the monotonicity of $a_n=1+frac{(-1)^n}{n}$, but what I'm getting isn't correct:
I just assume that $a_n$ is monotonically increasing, and if it isn't, I'll get something absurd:
So, $a_n$ is monotonically increasing iff
$$
begin{split}
a_{n+1} geq a_n
&iff left(1+frac{(-1)^{n+1}}{n+1}right)
- left(1+frac{(-1)^{n}}{n}right) geq 0 \
&iff frac{(-1)^{n+1}}{n+1} - frac{(-1)^{n}}{n} geq 0 \
&iff frac{(-1)^{n+1}}{n+1} geq frac{(-1)^{n}}{n},
end{split}
$$
Which doesn't make sense since $lim_{n to infty} frac{(-1)^n}{n}=0$ and $lim_{n to infty} frac{(-1)^{n+1}}{n+1}=0$, but the later one decreases more rapidly, so $a_n$ would be monotonically decreasing.
But, the solution says that if $n$ is even then $a_n$ is monotonically decreasing and if $n$ is odd then monotonically increasing. How do I prove that? What's wrong with what I did?
real-analysis calculus sequences-and-series monotone-functions
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
asked Dec 24 '18 at 15:51
E----E----
32410
$endgroup$
add a comment |
$begingroup$
I'm trying to study the monotonicity of $a_n=1+frac{(-1)^n}{n}$, but what I'm getting isn't correct:
I just assume that $a_n$ is monotonically increasing, and if it isn't, I'll get something absurd:
So, $a_n$ is monotonically increasing iff
$$
begin{split}
a_{n+1} geq a_n
&iff left(1+frac{(-1)^{n+1}}{n+1}right)
- left(1+frac{(-1)^{n}}{n}right) geq 0 \
&iff frac{(-1)^{n+1}}{n+1} - frac{(-1)^{n}}{n} geq 0 \
&iff frac{(-1)^{n+1}}{n+1} geq frac{(-1)^{n}}{n},
end{split}
$$
Which doesn't make sense since $lim_{n to infty} frac{(-1)^n}{n}=0$ and $lim_{n to infty} frac{(-1)^{n+1}}{n+1}=0$, but the later one decreases more rapidly, so $a_n$ would be monotonically decreasing.
But, the solution says that if $n$ is even then $a_n$ is monotonically decreasing and if $n$ is odd then monotonically increasing. How do I prove that? What's wrong with what I did?
real-analysis calculus sequences-and-series monotone-functions
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
asked Dec 24 '18 at 15:51
E----E----
32410
$endgroup$
I'm trying to study the monotonicity of $a_n=1+frac{(-1)^n}{n}$, but what I'm getting isn't correct:
I just assume that $a_n$ is monotonically increasing, and if it isn't, I'll get something absurd:
So, $a_n$ is monotonically increasing iff
$$
begin{split}
a_{n+1} geq a_n
&iff left(1+frac{(-1)^{n+1}}{n+1}right)
- left(1+frac{(-1)^{n}}{n}right) geq 0 \
&iff frac{(-1)^{n+1}}{n+1} - frac{(-1)^{n}}{n} geq 0 \
&iff frac{(-1)^{n+1}}{n+1} geq frac{(-1)^{n}}{n},
end{split}
$$
Which doesn't make sense since $lim_{n to infty} frac{(-1)^n}{n}=0$ and $lim_{n to infty} frac{(-1)^{n+1}}{n+1}=0$, but the later one decreases more rapidly, so $a_n$ would be monotonically decreasing.
But, the solution says that if $n$ is even then $a_n$ is monotonically decreasing and if $n$ is odd then monotonically increasing. How do I prove that? What's wrong with what I did?
real-analysis calculus sequences-and-series monotone-functions
real-analysis calculus sequences-and-series monotone-functions
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
asked Dec 24 '18 at 15:51
E----E----
32410
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
asked Dec 24 '18 at 15:51
E----E----
32410
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
edited Dec 24 '18 at 16:01
gt6989b
36.1k22557
36.1k22557
asked Dec 24 '18 at 15:51
E----E----
32410
asked Dec 24 '18 at 15:51
E----E----
32410
asked Dec 24 '18 at 15:51
E----E----
32410
32410
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
HINT
There is nothing wrong with your logic, but the last statement is false for any even $n$, since the RHS is positive and the LHS is negative.
To appreciate what is going on, graph the function:
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
$endgroup$
add a comment |
$begingroup$
Your sequence is oscillating around its limit point. If you take a look at two subsequences for odd and even indices, you'll notice that both of them are monotone. One is monotonically increasing, while the other is monotonically decreasing:
$$
a_{2k} = 1 + {1over 2k} \
a_{2k -1} = 1 - {1over 2k-1}
$$
But the whole sequence is not monotone.
Here is a visualization.
answered Dec 24 '18 at 15:56
romanroman
2,55921226
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add a comment |
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It is not monotonic since $a_{2n}>1$ and $a_{2n+1}<1$.
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
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add a comment |
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But it's not monotonically increasing because $a_{2n+1}=1-1/(2n+1)<1+1/2n=a_{2n}.$
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
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add a comment |
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The sequence $(a_n)_{ngeq 1}$ is not monotonic, but the subsequence $(a_{2k})_{kgeq 1}$ is decreasing and the subsequence $(a_{2k+1})_{kgeq 1}$ is increasing.
answered Dec 24 '18 at 15:56
MirceaMircea
1736
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
There is nothing wrong with your logic, but the last statement is false for any even $n$, since the RHS is positive and the LHS is negative.
To appreciate what is going on, graph the function:
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
$endgroup$
add a comment |
$begingroup$
HINT
There is nothing wrong with your logic, but the last statement is false for any even $n$, since the RHS is positive and the LHS is negative.
To appreciate what is going on, graph the function:
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
$endgroup$
add a comment |
$begingroup$
HINT
There is nothing wrong with your logic, but the last statement is false for any even $n$, since the RHS is positive and the LHS is negative.
To appreciate what is going on, graph the function:
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
$endgroup$
HINT
There is nothing wrong with your logic, but the last statement is false for any even $n$, since the RHS is positive and the LHS is negative.
To appreciate what is going on, graph the function:
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
answered Dec 24 '18 at 15:54
gt6989bgt6989b
36.1k22557
36.1k22557
add a comment |
add a comment |
$begingroup$
Your sequence is oscillating around its limit point. If you take a look at two subsequences for odd and even indices, you'll notice that both of them are monotone. One is monotonically increasing, while the other is monotonically decreasing:
$$
a_{2k} = 1 + {1over 2k} \
a_{2k -1} = 1 - {1over 2k-1}
$$
But the whole sequence is not monotone.
Here is a visualization.
answered Dec 24 '18 at 15:56
romanroman
2,55921226
$endgroup$
add a comment |
$begingroup$
Your sequence is oscillating around its limit point. If you take a look at two subsequences for odd and even indices, you'll notice that both of them are monotone. One is monotonically increasing, while the other is monotonically decreasing:
$$
a_{2k} = 1 + {1over 2k} \
a_{2k -1} = 1 - {1over 2k-1}
$$
But the whole sequence is not monotone.
Here is a visualization.
answered Dec 24 '18 at 15:56
romanroman
2,55921226
$endgroup$
add a comment |
$begingroup$
Your sequence is oscillating around its limit point. If you take a look at two subsequences for odd and even indices, you'll notice that both of them are monotone. One is monotonically increasing, while the other is monotonically decreasing:
$$
a_{2k} = 1 + {1over 2k} \
a_{2k -1} = 1 - {1over 2k-1}
$$
But the whole sequence is not monotone.
Here is a visualization.
answered Dec 24 '18 at 15:56
romanroman
2,55921226
$endgroup$
Your sequence is oscillating around its limit point. If you take a look at two subsequences for odd and even indices, you'll notice that both of them are monotone. One is monotonically increasing, while the other is monotonically decreasing:
$$
a_{2k} = 1 + {1over 2k} \
a_{2k -1} = 1 - {1over 2k-1}
$$
But the whole sequence is not monotone.
Here is a visualization.
answered Dec 24 '18 at 15:56
romanroman
2,55921226
edited Dec 24 '18 at 16:05
answered Dec 24 '18 at 15:56
romanroman
2,55921226
answered Dec 24 '18 at 15:56
romanroman
2,55921226
answered Dec 24 '18 at 15:56
romanroman
2,55921226
2,55921226
add a comment |
add a comment |
$begingroup$
It is not monotonic since $a_{2n}>1$ and $a_{2n+1}<1$.
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
$endgroup$
add a comment |
$begingroup$
It is not monotonic since $a_{2n}>1$ and $a_{2n+1}<1$.
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
$endgroup$
add a comment |
$begingroup$
It is not monotonic since $a_{2n}>1$ and $a_{2n+1}<1$.
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
$endgroup$
It is not monotonic since $a_{2n}>1$ and $a_{2n+1}<1$.
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
answered Dec 24 '18 at 16:37
TaladrisTaladris
4,92731933
4,92731933
add a comment |
add a comment |
$begingroup$
But it's not monotonically increasing because $a_{2n+1}=1-1/(2n+1)<1+1/2n=a_{2n}.$
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
$endgroup$
add a comment |
$begingroup$
But it's not monotonically increasing because $a_{2n+1}=1-1/(2n+1)<1+1/2n=a_{2n}.$
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
$endgroup$
add a comment |
$begingroup$
But it's not monotonically increasing because $a_{2n+1}=1-1/(2n+1)<1+1/2n=a_{2n}.$
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
$endgroup$
But it's not monotonically increasing because $a_{2n+1}=1-1/(2n+1)<1+1/2n=a_{2n}.$
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
answered Dec 24 '18 at 15:55
John_WickJohn_Wick
1,626111
1,626111
add a comment |
add a comment |
$begingroup$
The sequence $(a_n)_{ngeq 1}$ is not monotonic, but the subsequence $(a_{2k})_{kgeq 1}$ is decreasing and the subsequence $(a_{2k+1})_{kgeq 1}$ is increasing.
answered Dec 24 '18 at 15:56
MirceaMircea
1736
$endgroup$
add a comment |
$begingroup$
The sequence $(a_n)_{ngeq 1}$ is not monotonic, but the subsequence $(a_{2k})_{kgeq 1}$ is decreasing and the subsequence $(a_{2k+1})_{kgeq 1}$ is increasing.
answered Dec 24 '18 at 15:56
MirceaMircea
1736
$endgroup$
add a comment |
$begingroup$
The sequence $(a_n)_{ngeq 1}$ is not monotonic, but the subsequence $(a_{2k})_{kgeq 1}$ is decreasing and the subsequence $(a_{2k+1})_{kgeq 1}$ is increasing.
answered Dec 24 '18 at 15:56
MirceaMircea
1736
$endgroup$
The sequence $(a_n)_{ngeq 1}$ is not monotonic, but the subsequence $(a_{2k})_{kgeq 1}$ is decreasing and the subsequence $(a_{2k+1})_{kgeq 1}$ is increasing.
answered Dec 24 '18 at 15:56
MirceaMircea
1736
answered Dec 24 '18 at 15:56
MirceaMircea
1736
answered Dec 24 '18 at 15:56
MirceaMircea
1736
answered Dec 24 '18 at 15:56
MirceaMircea
1736
1736
add a comment |
add a comment |
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