Mixing
Proving that sequence is decreasing
$begingroup$
I was looking at the first answer to this question Simple proof of showing the Harmonic number $H_n = Theta (log n)$, which claims that the following sequence is decreasing:
$$
f(n) = H_n - log(n)
$$
where $log$ is the natural log and $H_n = sum_{i = 1}^n frac{1}{i}$.
Here is my attempt to verify this:
Observe that
$$
f(n + 1) - f(n) = H_{n + 1} - log(n + 1) - (H_n - log(n)) = frac{1}{n + 1} + log left(frac{n}{n + 1} right).
$$
But this last term is positive, so it appears that the sequence is increasing! Where is my mistake?
real-analysis
asked Jan 22 at 1:14
SubmarineSubmarine
61
$endgroup$
|
show 1 more comment
$begingroup$
I was looking at the first answer to this question Simple proof of showing the Harmonic number $H_n = Theta (log n)$, which claims that the following sequence is decreasing:
$$
f(n) = H_n - log(n)
$$
where $log$ is the natural log and $H_n = sum_{i = 1}^n frac{1}{i}$.
Here is my attempt to verify this:
Observe that
$$
f(n + 1) - f(n) = H_{n + 1} - log(n + 1) - (H_n - log(n)) = frac{1}{n + 1} + log left(frac{n}{n + 1} right).
$$
But this last term is positive, so it appears that the sequence is increasing! Where is my mistake?
real-analysis
asked Jan 22 at 1:14
SubmarineSubmarine
61
$endgroup$
3
$begingroup$
Welcome to MSE. Note that although $frac{1}{n + 1}$ is positive, $log left(frac{n}{n+1}right)$ is negative as $frac{n}{n+1} lt 1$.
$endgroup$
– John Omielan
Jan 22 at 1:16
3
$begingroup$
no, $n/(n+1)$ is smaller than one, so its logarithm is negative.
$endgroup$
– Will Jagy
Jan 22 at 1:17
$begingroup$
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:18
$begingroup$
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion.
$endgroup$
– John Omielan
Jan 22 at 1:24
$begingroup$
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:26
|
show 1 more comment
$begingroup$
I was looking at the first answer to this question Simple proof of showing the Harmonic number $H_n = Theta (log n)$, which claims that the following sequence is decreasing:
$$
f(n) = H_n - log(n)
$$
where $log$ is the natural log and $H_n = sum_{i = 1}^n frac{1}{i}$.
Here is my attempt to verify this:
Observe that
$$
f(n + 1) - f(n) = H_{n + 1} - log(n + 1) - (H_n - log(n)) = frac{1}{n + 1} + log left(frac{n}{n + 1} right).
$$
But this last term is positive, so it appears that the sequence is increasing! Where is my mistake?
real-analysis
asked Jan 22 at 1:14
SubmarineSubmarine
61
$endgroup$
I was looking at the first answer to this question Simple proof of showing the Harmonic number $H_n = Theta (log n)$, which claims that the following sequence is decreasing:
$$
f(n) = H_n - log(n)
$$
where $log$ is the natural log and $H_n = sum_{i = 1}^n frac{1}{i}$.
Here is my attempt to verify this:
Observe that
$$
f(n + 1) - f(n) = H_{n + 1} - log(n + 1) - (H_n - log(n)) = frac{1}{n + 1} + log left(frac{n}{n + 1} right).
$$
But this last term is positive, so it appears that the sequence is increasing! Where is my mistake?
real-analysis
real-analysis
asked Jan 22 at 1:14
SubmarineSubmarine
61
asked Jan 22 at 1:14
SubmarineSubmarine
61
asked Jan 22 at 1:14
SubmarineSubmarine
61
asked Jan 22 at 1:14
SubmarineSubmarine
61
asked Jan 22 at 1:14
SubmarineSubmarine
61
61
3
$begingroup$
Welcome to MSE. Note that although $frac{1}{n + 1}$ is positive, $log left(frac{n}{n+1}right)$ is negative as $frac{n}{n+1} lt 1$.
$endgroup$
– John Omielan
Jan 22 at 1:16
3
$begingroup$
no, $n/(n+1)$ is smaller than one, so its logarithm is negative.
$endgroup$
– Will Jagy
Jan 22 at 1:17
$begingroup$
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:18
$begingroup$
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion.
$endgroup$
– John Omielan
Jan 22 at 1:24
$begingroup$
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:26
|
show 1 more comment
3
$begingroup$
Welcome to MSE. Note that although $frac{1}{n + 1}$ is positive, $log left(frac{n}{n+1}right)$ is negative as $frac{n}{n+1} lt 1$.
$endgroup$
– John Omielan
Jan 22 at 1:16
3
$begingroup$
no, $n/(n+1)$ is smaller than one, so its logarithm is negative.
$endgroup$
– Will Jagy
Jan 22 at 1:17
$begingroup$
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:18
$begingroup$
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion.
$endgroup$
– John Omielan
Jan 22 at 1:24
$begingroup$
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:26
3
3
$begingroup$
Welcome to MSE. Note that although $frac{1}{n + 1}$ is positive, $log left(frac{n}{n+1}right)$ is negative as $frac{n}{n+1} lt 1$.
$endgroup$
– John Omielan
Jan 22 at 1:16
$begingroup$
Welcome to MSE. Note that although $frac{1}{n + 1}$ is positive, $log left(frac{n}{n+1}right)$ is negative as $frac{n}{n+1} lt 1$.
$endgroup$
– John Omielan
Jan 22 at 1:16
3
3
$begingroup$
no, $n/(n+1)$ is smaller than one, so its logarithm is negative.
$endgroup$
– Will Jagy
Jan 22 at 1:17
$begingroup$
no, $n/(n+1)$ is smaller than one, so its logarithm is negative.
$endgroup$
– Will Jagy
Jan 22 at 1:17
$begingroup$
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:18
$begingroup$
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:18
$begingroup$
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion.
$endgroup$
– John Omielan
Jan 22 at 1:24
$begingroup$
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion.
$endgroup$
– John Omielan
Jan 22 at 1:24
$begingroup$
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:26
$begingroup$
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:26
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$$frac{1}{n+1} < int_n^{n+1} frac{dt}{t} = log (n+1) - log n \ implies frac{1}{n+1} -log (n+1) < - log n$$
answered Jan 22 at 1:29
RRLRRL
52.4k42573
$endgroup$
add a comment |
$begingroup$
As the other comments noted, although $frac{n}{n+1}$ is positive, it is also less than 1 so its natural logarithm must be negative.
Thus, the problem comes down to showing that
$$frac{1}{x} + lnBig(frac{x-1}{x}Big)<0$$
That is equivalent to
$$f(x)=frac{1}{x} - lnBig(frac{x}{x-1}Big) < 0.$$
It's easy to see that
$$lim_{x to infty} f(x) = 0$$
and also that
$$f^prime (x) = frac{1}{x^2(x-1)},$$
which is positive for all $x<1$. Thus, $f$ is a strictly inscreasing function. As such, $max(f(x))=lim_{xtoinfty}f(x)=0$.
Additionally, $f$ is not defined at $x=0$ so $f$ must be strictly negative.
answered Jan 22 at 2:00
LorenzoLorenzo
1011
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
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$begingroup$
Hint:
$$frac{1}{n+1} < int_n^{n+1} frac{dt}{t} = log (n+1) - log n \ implies frac{1}{n+1} -log (n+1) < - log n$$
answered Jan 22 at 1:29
RRLRRL
52.4k42573
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac{1}{n+1} < int_n^{n+1} frac{dt}{t} = log (n+1) - log n \ implies frac{1}{n+1} -log (n+1) < - log n$$
answered Jan 22 at 1:29
RRLRRL
52.4k42573
$endgroup$
add a comment |
$begingroup$
Hint:
$$frac{1}{n+1} < int_n^{n+1} frac{dt}{t} = log (n+1) - log n \ implies frac{1}{n+1} -log (n+1) < - log n$$
answered Jan 22 at 1:29
RRLRRL
52.4k42573
$endgroup$
Hint:
$$frac{1}{n+1} < int_n^{n+1} frac{dt}{t} = log (n+1) - log n \ implies frac{1}{n+1} -log (n+1) < - log n$$
answered Jan 22 at 1:29
RRLRRL
52.4k42573
answered Jan 22 at 1:29
RRLRRL
52.4k42573
answered Jan 22 at 1:29
RRLRRL
52.4k42573
answered Jan 22 at 1:29
RRLRRL
52.4k42573
52.4k42573
add a comment |
add a comment |
$begingroup$
As the other comments noted, although $frac{n}{n+1}$ is positive, it is also less than 1 so its natural logarithm must be negative.
Thus, the problem comes down to showing that
$$frac{1}{x} + lnBig(frac{x-1}{x}Big)<0$$
That is equivalent to
$$f(x)=frac{1}{x} - lnBig(frac{x}{x-1}Big) < 0.$$
It's easy to see that
$$lim_{x to infty} f(x) = 0$$
and also that
$$f^prime (x) = frac{1}{x^2(x-1)},$$
which is positive for all $x<1$. Thus, $f$ is a strictly inscreasing function. As such, $max(f(x))=lim_{xtoinfty}f(x)=0$.
Additionally, $f$ is not defined at $x=0$ so $f$ must be strictly negative.
answered Jan 22 at 2:00
LorenzoLorenzo
1011
$endgroup$
add a comment |
$begingroup$
As the other comments noted, although $frac{n}{n+1}$ is positive, it is also less than 1 so its natural logarithm must be negative.
Thus, the problem comes down to showing that
$$frac{1}{x} + lnBig(frac{x-1}{x}Big)<0$$
That is equivalent to
$$f(x)=frac{1}{x} - lnBig(frac{x}{x-1}Big) < 0.$$
It's easy to see that
$$lim_{x to infty} f(x) = 0$$
and also that
$$f^prime (x) = frac{1}{x^2(x-1)},$$
which is positive for all $x<1$. Thus, $f$ is a strictly inscreasing function. As such, $max(f(x))=lim_{xtoinfty}f(x)=0$.
Additionally, $f$ is not defined at $x=0$ so $f$ must be strictly negative.
answered Jan 22 at 2:00
LorenzoLorenzo
1011
$endgroup$
add a comment |
$begingroup$
As the other comments noted, although $frac{n}{n+1}$ is positive, it is also less than 1 so its natural logarithm must be negative.
Thus, the problem comes down to showing that
$$frac{1}{x} + lnBig(frac{x-1}{x}Big)<0$$
That is equivalent to
$$f(x)=frac{1}{x} - lnBig(frac{x}{x-1}Big) < 0.$$
It's easy to see that
$$lim_{x to infty} f(x) = 0$$
and also that
$$f^prime (x) = frac{1}{x^2(x-1)},$$
which is positive for all $x<1$. Thus, $f$ is a strictly inscreasing function. As such, $max(f(x))=lim_{xtoinfty}f(x)=0$.
Additionally, $f$ is not defined at $x=0$ so $f$ must be strictly negative.
answered Jan 22 at 2:00
LorenzoLorenzo
1011
$endgroup$
As the other comments noted, although $frac{n}{n+1}$ is positive, it is also less than 1 so its natural logarithm must be negative.
Thus, the problem comes down to showing that
$$frac{1}{x} + lnBig(frac{x-1}{x}Big)<0$$
That is equivalent to
$$f(x)=frac{1}{x} - lnBig(frac{x}{x-1}Big) < 0.$$
It's easy to see that
$$lim_{x to infty} f(x) = 0$$
and also that
$$f^prime (x) = frac{1}{x^2(x-1)},$$
which is positive for all $x<1$. Thus, $f$ is a strictly inscreasing function. As such, $max(f(x))=lim_{xtoinfty}f(x)=0$.
Additionally, $f$ is not defined at $x=0$ so $f$ must be strictly negative.
answered Jan 22 at 2:00
LorenzoLorenzo
1011
answered Jan 22 at 2:00
LorenzoLorenzo
1011
answered Jan 22 at 2:00
LorenzoLorenzo
1011
answered Jan 22 at 2:00
LorenzoLorenzo
1011
1011
add a comment |
add a comment |
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3
$begingroup$
Welcome to MSE. Note that although $frac{1}{n + 1}$ is positive, $log left(frac{n}{n+1}right)$ is negative as $frac{n}{n+1} lt 1$.
$endgroup$
– John Omielan
Jan 22 at 1:16
3
$begingroup$
no, $n/(n+1)$ is smaller than one, so its logarithm is negative.
$endgroup$
– Will Jagy
Jan 22 at 1:17
$begingroup$
@JohnOmielan Thank you. What would be a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:18
$begingroup$
@Submarine I'm not sure what a particularly "good" way to show this inequality is negative, with it depending to a certain extent on what you have learned & know how to use. However, I would use a Taylor expansion of the natural log, in particular, the first one in Taylor Series Expansions of Logarimathic Functions and show the remaining terms are negative. However, this is something I'm not particularly expert at, so perhaps somebody here can give you a better suggestion.
$endgroup$
– John Omielan
Jan 22 at 1:24
$begingroup$
@WillJagy Thank you for the comment. Do you know a good way to show that this inequality is negative?
$endgroup$
– Submarine
Jan 22 at 1:26