Mixing
Proof of harmonic series divergence [closed]
$begingroup$
Prove that ∑_(k=1)^∞ 1/k is divergent
Proof
Could anyone possibly help explain my professor’s notation in this proof? Struggling to understand where S_(2^n-1)≥n/2 came from as well as the manipulation of the sums that follow.
Any help is greatly appreciated
sequences-and-series analysis divergence
asked Jan 8 at 11:56
frenny13frenny13
1
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closed as off-topic by amWhy, Paul Frost, RRL, Alexander Gruber♦ Jan 9 at 0:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, RRL, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that ∑_(k=1)^∞ 1/k is divergent
Proof
Could anyone possibly help explain my professor’s notation in this proof? Struggling to understand where S_(2^n-1)≥n/2 came from as well as the manipulation of the sums that follow.
Any help is greatly appreciated
sequences-and-series analysis divergence
asked Jan 8 at 11:56
frenny13frenny13
1
$endgroup$
closed as off-topic by amWhy, Paul Frost, RRL, Alexander Gruber♦ Jan 9 at 0:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, RRL, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that ∑_(k=1)^∞ 1/k is divergent
Proof
Could anyone possibly help explain my professor’s notation in this proof? Struggling to understand where S_(2^n-1)≥n/2 came from as well as the manipulation of the sums that follow.
Any help is greatly appreciated
sequences-and-series analysis divergence
asked Jan 8 at 11:56
frenny13frenny13
1
$endgroup$
Prove that ∑_(k=1)^∞ 1/k is divergent
Proof
Could anyone possibly help explain my professor’s notation in this proof? Struggling to understand where S_(2^n-1)≥n/2 came from as well as the manipulation of the sums that follow.
Any help is greatly appreciated
sequences-and-series analysis divergence
sequences-and-series analysis divergence
asked Jan 8 at 11:56
frenny13frenny13
1
asked Jan 8 at 11:56
frenny13frenny13
1
asked Jan 8 at 11:56
frenny13frenny13
1
asked Jan 8 at 11:56
frenny13frenny13
1
asked Jan 8 at 11:56
frenny13frenny13
1
1
closed as off-topic by amWhy, Paul Frost, RRL, Alexander Gruber♦ Jan 9 at 0:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, RRL, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Paul Frost, RRL, Alexander Gruber♦ Jan 9 at 0:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, RRL, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Ben W gave you an answer. I will give you a very simple proof for the divergence of the harmonic series:
Let $S_n :=1+frac{1}{2} + ....+ frac{1}{n} $. Then
$$(*) quad S_{2n}= S_n +frac{1}{n+1} +...+ frac{1}{2n} ge S_n +frac{1}{2n} +...+ frac{1}{2n}= S_n +frac{1}{2}.$$
Now assume that $(S_n)$ is convergent, with limit $S$. From $(*)$ we get $S ge S+frac{1}{2}$, a contradiction.
answered Jan 8 at 12:15
FredFred
45.3k1847
$endgroup$
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
add a comment |
$begingroup$
Any sum of the form
$$sum_{i=1}^{2^n-1}f(i)$$
can be written as the double sum
$$sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}f(i)$$
To see this, consider a concrete example $n=3$. Then $2^n-1=7$ so that
$$sum_{i=1}^{2^3-1}=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$
$$=[f(1)]+[f(2)+f(3)]+[f(4)+f(5)+f(6)+f(7)]$$
$$=sum_{i=2^{1-1}}^{2^1-1}f(i)+sum_{i=2^{2-1}}^{2^2-1}f(i)+sum_{i=2^{3-1}}^{2^3-1}f(i)$$
$$=sum_{k=1}^3sum_{i=2^{k-1}}^{2^k-1}f(i)$$
See it now? Hence we can write
$$S_{2^n-1}=sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}=1+sum_{k=2}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}$$
and the rest should be clear.
P.S. Actually, splitting off the "1" is entirely unnecessary, although not incorrect.
answered Jan 8 at 12:06
Ben WBen W
2,274615
$endgroup$
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Ben W gave you an answer. I will give you a very simple proof for the divergence of the harmonic series:
Let $S_n :=1+frac{1}{2} + ....+ frac{1}{n} $. Then
$$(*) quad S_{2n}= S_n +frac{1}{n+1} +...+ frac{1}{2n} ge S_n +frac{1}{2n} +...+ frac{1}{2n}= S_n +frac{1}{2}.$$
Now assume that $(S_n)$ is convergent, with limit $S$. From $(*)$ we get $S ge S+frac{1}{2}$, a contradiction.
answered Jan 8 at 12:15
FredFred
45.3k1847
$endgroup$
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
add a comment |
$begingroup$
Ben W gave you an answer. I will give you a very simple proof for the divergence of the harmonic series:
Let $S_n :=1+frac{1}{2} + ....+ frac{1}{n} $. Then
$$(*) quad S_{2n}= S_n +frac{1}{n+1} +...+ frac{1}{2n} ge S_n +frac{1}{2n} +...+ frac{1}{2n}= S_n +frac{1}{2}.$$
Now assume that $(S_n)$ is convergent, with limit $S$. From $(*)$ we get $S ge S+frac{1}{2}$, a contradiction.
answered Jan 8 at 12:15
FredFred
45.3k1847
$endgroup$
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
add a comment |
$begingroup$
Ben W gave you an answer. I will give you a very simple proof for the divergence of the harmonic series:
Let $S_n :=1+frac{1}{2} + ....+ frac{1}{n} $. Then
$$(*) quad S_{2n}= S_n +frac{1}{n+1} +...+ frac{1}{2n} ge S_n +frac{1}{2n} +...+ frac{1}{2n}= S_n +frac{1}{2}.$$
Now assume that $(S_n)$ is convergent, with limit $S$. From $(*)$ we get $S ge S+frac{1}{2}$, a contradiction.
answered Jan 8 at 12:15
FredFred
45.3k1847
$endgroup$
Ben W gave you an answer. I will give you a very simple proof for the divergence of the harmonic series:
Let $S_n :=1+frac{1}{2} + ....+ frac{1}{n} $. Then
$$(*) quad S_{2n}= S_n +frac{1}{n+1} +...+ frac{1}{2n} ge S_n +frac{1}{2n} +...+ frac{1}{2n}= S_n +frac{1}{2}.$$
Now assume that $(S_n)$ is convergent, with limit $S$. From $(*)$ we get $S ge S+frac{1}{2}$, a contradiction.
answered Jan 8 at 12:15
FredFred
45.3k1847
answered Jan 8 at 12:15
FredFred
45.3k1847
answered Jan 8 at 12:15
FredFred
45.3k1847
answered Jan 8 at 12:15
FredFred
45.3k1847
45.3k1847
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
add a comment |
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
$begingroup$
Yeah I was trying to help him understand his professor's much longer proof, but this is quite nice!
$endgroup$
– Ben W
Jan 8 at 12:17
add a comment |
$begingroup$
Any sum of the form
$$sum_{i=1}^{2^n-1}f(i)$$
can be written as the double sum
$$sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}f(i)$$
To see this, consider a concrete example $n=3$. Then $2^n-1=7$ so that
$$sum_{i=1}^{2^3-1}=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$
$$=[f(1)]+[f(2)+f(3)]+[f(4)+f(5)+f(6)+f(7)]$$
$$=sum_{i=2^{1-1}}^{2^1-1}f(i)+sum_{i=2^{2-1}}^{2^2-1}f(i)+sum_{i=2^{3-1}}^{2^3-1}f(i)$$
$$=sum_{k=1}^3sum_{i=2^{k-1}}^{2^k-1}f(i)$$
See it now? Hence we can write
$$S_{2^n-1}=sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}=1+sum_{k=2}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}$$
and the rest should be clear.
P.S. Actually, splitting off the "1" is entirely unnecessary, although not incorrect.
answered Jan 8 at 12:06
Ben WBen W
2,274615
$endgroup$
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
add a comment |
$begingroup$
Any sum of the form
$$sum_{i=1}^{2^n-1}f(i)$$
can be written as the double sum
$$sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}f(i)$$
To see this, consider a concrete example $n=3$. Then $2^n-1=7$ so that
$$sum_{i=1}^{2^3-1}=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$
$$=[f(1)]+[f(2)+f(3)]+[f(4)+f(5)+f(6)+f(7)]$$
$$=sum_{i=2^{1-1}}^{2^1-1}f(i)+sum_{i=2^{2-1}}^{2^2-1}f(i)+sum_{i=2^{3-1}}^{2^3-1}f(i)$$
$$=sum_{k=1}^3sum_{i=2^{k-1}}^{2^k-1}f(i)$$
See it now? Hence we can write
$$S_{2^n-1}=sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}=1+sum_{k=2}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}$$
and the rest should be clear.
P.S. Actually, splitting off the "1" is entirely unnecessary, although not incorrect.
answered Jan 8 at 12:06
Ben WBen W
2,274615
$endgroup$
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
add a comment |
$begingroup$
Any sum of the form
$$sum_{i=1}^{2^n-1}f(i)$$
can be written as the double sum
$$sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}f(i)$$
To see this, consider a concrete example $n=3$. Then $2^n-1=7$ so that
$$sum_{i=1}^{2^3-1}=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$
$$=[f(1)]+[f(2)+f(3)]+[f(4)+f(5)+f(6)+f(7)]$$
$$=sum_{i=2^{1-1}}^{2^1-1}f(i)+sum_{i=2^{2-1}}^{2^2-1}f(i)+sum_{i=2^{3-1}}^{2^3-1}f(i)$$
$$=sum_{k=1}^3sum_{i=2^{k-1}}^{2^k-1}f(i)$$
See it now? Hence we can write
$$S_{2^n-1}=sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}=1+sum_{k=2}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}$$
and the rest should be clear.
P.S. Actually, splitting off the "1" is entirely unnecessary, although not incorrect.
answered Jan 8 at 12:06
Ben WBen W
2,274615
$endgroup$
Any sum of the form
$$sum_{i=1}^{2^n-1}f(i)$$
can be written as the double sum
$$sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}f(i)$$
To see this, consider a concrete example $n=3$. Then $2^n-1=7$ so that
$$sum_{i=1}^{2^3-1}=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)$$
$$=[f(1)]+[f(2)+f(3)]+[f(4)+f(5)+f(6)+f(7)]$$
$$=sum_{i=2^{1-1}}^{2^1-1}f(i)+sum_{i=2^{2-1}}^{2^2-1}f(i)+sum_{i=2^{3-1}}^{2^3-1}f(i)$$
$$=sum_{k=1}^3sum_{i=2^{k-1}}^{2^k-1}f(i)$$
See it now? Hence we can write
$$S_{2^n-1}=sum_{k=1}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}=1+sum_{k=2}^nsum_{i=2^{k-1}}^{2^k-1}frac{1}{i}$$
and the rest should be clear.
P.S. Actually, splitting off the "1" is entirely unnecessary, although not incorrect.
answered Jan 8 at 12:06
Ben WBen W
2,274615
answered Jan 8 at 12:06
Ben WBen W
2,274615
answered Jan 8 at 12:06
Ben WBen W
2,274615
answered Jan 8 at 12:06
Ben WBen W
2,274615
2,274615
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
add a comment |
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
Great thanks for explaining that it makes sense now. How come S_(2^n-1)≥n/2 in the first sentence of the original proof was chosen as a sufficient to show it is unbounded and divergent? What motivated that specific inequality. Cheers
$endgroup$
– frenny13
Jan 8 at 12:12
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
$begingroup$
It was simply convenient. Since $n/2toinfty$ with $n/2leq S_{2^n-1}$, we get $S_{2^n-1}toinfty$ as well. But any expression that goes to $infty$ will do.
$endgroup$
– Ben W
Jan 8 at 12:13
add a comment |
