Mixing
Divergent infinite series $n!e^n/n^n$ - simpler proof of divergence?
$begingroup$
$$ sum_{n=1}^infty frac{n!e^n}{n^n} $$
Where $e$ is Euler's number.
Recently on calculus class we were covering convergence tests and my group got stuck with this infinite series. Our calculus teacher told us that it diverges but he doesn't see any elementary way or simple test to prove it. He told us something about Euler's Gamma function and said that he knows it diverges because of Stirling's formula but this is not quite our level yet.
I tried to think about it myself but I established only that this is some kind of an edge case.
We could generalize the problem to series of $frac{n!a^n}{n^n}$.
Then, from the ratio test we know that it converges for $a <e$ and diverges for $a>e$. For $a=e$ the test is inconclusive.
I've also tried to compare it with various divergent series, but nothing worked...
Does anybody have any idea for a simple solution?
real-analysis calculus analysis convergence divergent-series
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
asked Jan 11 at 11:16
SkyfallSkyfall
234
$endgroup$
add a comment |
$begingroup$
$$ sum_{n=1}^infty frac{n!e^n}{n^n} $$
Where $e$ is Euler's number.
Recently on calculus class we were covering convergence tests and my group got stuck with this infinite series. Our calculus teacher told us that it diverges but he doesn't see any elementary way or simple test to prove it. He told us something about Euler's Gamma function and said that he knows it diverges because of Stirling's formula but this is not quite our level yet.
I tried to think about it myself but I established only that this is some kind of an edge case.
We could generalize the problem to series of $frac{n!a^n}{n^n}$.
Then, from the ratio test we know that it converges for $a <e$ and diverges for $a>e$. For $a=e$ the test is inconclusive.
I've also tried to compare it with various divergent series, but nothing worked...
Does anybody have any idea for a simple solution?
real-analysis calculus analysis convergence divergent-series
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
asked Jan 11 at 11:16
SkyfallSkyfall
234
$endgroup$
$begingroup$
It would be sufficient to show that the limit of the expression for $nrightarrow infty$ is not $0$, but I have no simple idea yet.
$endgroup$
– Peter
Jan 11 at 11:20
$begingroup$
Maybe you can use the Stirling's approximation
$endgroup$
– Ixion
Jan 11 at 11:27
$begingroup$
Hi and welcome to the M.SE. If you want to add beautifully looking and mathematical text to your questions/answers, you may want to have a look at the Math Jax quick reference: you already did a good job by providing context to your problem.
$endgroup$
– Daniele Tampieri
Jan 11 at 11:27
2
$begingroup$
@Ixion "he knows it diverges because of Stirling's formula but this is not quite our level yet."
$endgroup$
– Wojowu
Jan 11 at 11:28
$begingroup$
@Wojowu, ouch, thanks... My bad!
$endgroup$
– Ixion
Jan 11 at 11:29
add a comment |
$begingroup$
$$ sum_{n=1}^infty frac{n!e^n}{n^n} $$
Where $e$ is Euler's number.
Recently on calculus class we were covering convergence tests and my group got stuck with this infinite series. Our calculus teacher told us that it diverges but he doesn't see any elementary way or simple test to prove it. He told us something about Euler's Gamma function and said that he knows it diverges because of Stirling's formula but this is not quite our level yet.
I tried to think about it myself but I established only that this is some kind of an edge case.
We could generalize the problem to series of $frac{n!a^n}{n^n}$.
Then, from the ratio test we know that it converges for $a <e$ and diverges for $a>e$. For $a=e$ the test is inconclusive.
I've also tried to compare it with various divergent series, but nothing worked...
Does anybody have any idea for a simple solution?
real-analysis calculus analysis convergence divergent-series
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
asked Jan 11 at 11:16
SkyfallSkyfall
234
$endgroup$
$$ sum_{n=1}^infty frac{n!e^n}{n^n} $$
Where $e$ is Euler's number.
Recently on calculus class we were covering convergence tests and my group got stuck with this infinite series. Our calculus teacher told us that it diverges but he doesn't see any elementary way or simple test to prove it. He told us something about Euler's Gamma function and said that he knows it diverges because of Stirling's formula but this is not quite our level yet.
I tried to think about it myself but I established only that this is some kind of an edge case.
We could generalize the problem to series of $frac{n!a^n}{n^n}$.
Then, from the ratio test we know that it converges for $a <e$ and diverges for $a>e$. For $a=e$ the test is inconclusive.
I've also tried to compare it with various divergent series, but nothing worked...
Does anybody have any idea for a simple solution?
real-analysis calculus analysis convergence divergent-series
real-analysis calculus analysis convergence divergent-series
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
asked Jan 11 at 11:16
SkyfallSkyfall
234
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
asked Jan 11 at 11:16
SkyfallSkyfall
234
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
edited Jan 11 at 11:27
José Carlos Santos
160k22126232
160k22126232
asked Jan 11 at 11:16
SkyfallSkyfall
234
asked Jan 11 at 11:16
SkyfallSkyfall
234
asked Jan 11 at 11:16
SkyfallSkyfall
234
234
$begingroup$
It would be sufficient to show that the limit of the expression for $nrightarrow infty$ is not $0$, but I have no simple idea yet.
$endgroup$
– Peter
Jan 11 at 11:20
$begingroup$
Maybe you can use the Stirling's approximation
$endgroup$
– Ixion
Jan 11 at 11:27
$begingroup$
Hi and welcome to the M.SE. If you want to add beautifully looking and mathematical text to your questions/answers, you may want to have a look at the Math Jax quick reference: you already did a good job by providing context to your problem.
$endgroup$
– Daniele Tampieri
Jan 11 at 11:27
2
$begingroup$
@Ixion "he knows it diverges because of Stirling's formula but this is not quite our level yet."
$endgroup$
– Wojowu
Jan 11 at 11:28
$begingroup$
@Wojowu, ouch, thanks... My bad!
$endgroup$
– Ixion
Jan 11 at 11:29
add a comment |
$begingroup$
It would be sufficient to show that the limit of the expression for $nrightarrow infty$ is not $0$, but I have no simple idea yet.
$endgroup$
– Peter
Jan 11 at 11:20
$begingroup$
Maybe you can use the Stirling's approximation
$endgroup$
– Ixion
Jan 11 at 11:27
$begingroup$
Hi and welcome to the M.SE. If you want to add beautifully looking and mathematical text to your questions/answers, you may want to have a look at the Math Jax quick reference: you already did a good job by providing context to your problem.
$endgroup$
– Daniele Tampieri
Jan 11 at 11:27
2
$begingroup$
@Ixion "he knows it diverges because of Stirling's formula but this is not quite our level yet."
$endgroup$
– Wojowu
Jan 11 at 11:28
$begingroup$
@Wojowu, ouch, thanks... My bad!
$endgroup$
– Ixion
Jan 11 at 11:29
$begingroup$
It would be sufficient to show that the limit of the expression for $nrightarrow infty$ is not $0$, but I have no simple idea yet.
$endgroup$
– Peter
Jan 11 at 11:20
$begingroup$
It would be sufficient to show that the limit of the expression for $nrightarrow infty$ is not $0$, but I have no simple idea yet.
$endgroup$
– Peter
Jan 11 at 11:20
$begingroup$
Maybe you can use the Stirling's approximation
$endgroup$
– Ixion
Jan 11 at 11:27
$begingroup$
Maybe you can use the Stirling's approximation
$endgroup$
– Ixion
Jan 11 at 11:27
$begingroup$
Hi and welcome to the M.SE. If you want to add beautifully looking and mathematical text to your questions/answers, you may want to have a look at the Math Jax quick reference: you already did a good job by providing context to your problem.
$endgroup$
– Daniele Tampieri
Jan 11 at 11:27
$begingroup$
Hi and welcome to the M.SE. If you want to add beautifully looking and mathematical text to your questions/answers, you may want to have a look at the Math Jax quick reference: you already did a good job by providing context to your problem.
$endgroup$
– Daniele Tampieri
Jan 11 at 11:27
2
2
$begingroup$
@Ixion "he knows it diverges because of Stirling's formula but this is not quite our level yet."
$endgroup$
– Wojowu
Jan 11 at 11:28
$begingroup$
@Ixion "he knows it diverges because of Stirling's formula but this is not quite our level yet."
$endgroup$
– Wojowu
Jan 11 at 11:28
$begingroup$
@Wojowu, ouch, thanks... My bad!
$endgroup$
– Ixion
Jan 11 at 11:29
$begingroup$
@Wojowu, ouch, thanks... My bad!
$endgroup$
– Ixion
Jan 11 at 11:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note
$$frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=frac{(n+1)e}{(n+1)^{n+1}/n^n}=frac{e}{(n+1)^n/n}=frac{e}{(1+1/n)^n}>1$$
since $e>(1+1/n)^n$ for all $n$. It follows that
$$frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>frac{n!e^n}{n^n},$$
so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.
Now, perhaps your teacher meant $frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $ncdotfrac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $frac{n^n}{n!e^n}geqfrac{1}{en}$, and by divergence of harmonic series this series diverges too.
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
$endgroup$
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
add a comment |
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$begingroup$
Note
$$frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=frac{(n+1)e}{(n+1)^{n+1}/n^n}=frac{e}{(n+1)^n/n}=frac{e}{(1+1/n)^n}>1$$
since $e>(1+1/n)^n$ for all $n$. It follows that
$$frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>frac{n!e^n}{n^n},$$
so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.
Now, perhaps your teacher meant $frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $ncdotfrac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $frac{n^n}{n!e^n}geqfrac{1}{en}$, and by divergence of harmonic series this series diverges too.
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
$endgroup$
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
add a comment |
$begingroup$
Note
$$frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=frac{(n+1)e}{(n+1)^{n+1}/n^n}=frac{e}{(n+1)^n/n}=frac{e}{(1+1/n)^n}>1$$
since $e>(1+1/n)^n$ for all $n$. It follows that
$$frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>frac{n!e^n}{n^n},$$
so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.
Now, perhaps your teacher meant $frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $ncdotfrac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $frac{n^n}{n!e^n}geqfrac{1}{en}$, and by divergence of harmonic series this series diverges too.
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
$endgroup$
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
add a comment |
$begingroup$
Note
$$frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=frac{(n+1)e}{(n+1)^{n+1}/n^n}=frac{e}{(n+1)^n/n}=frac{e}{(1+1/n)^n}>1$$
since $e>(1+1/n)^n$ for all $n$. It follows that
$$frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>frac{n!e^n}{n^n},$$
so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.
Now, perhaps your teacher meant $frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $ncdotfrac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $frac{n^n}{n!e^n}geqfrac{1}{en}$, and by divergence of harmonic series this series diverges too.
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
$endgroup$
Note
$$frac{(n+1)!e^{n+1}/(n+1)^{n+1}}{n!e^n/n^n}=frac{(n+1)e}{(n+1)^{n+1}/n^n}=frac{e}{(n+1)^n/n}=frac{e}{(1+1/n)^n}>1$$
since $e>(1+1/n)^n$ for all $n$. It follows that
$$frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}>frac{n!e^n}{n^n},$$
so the terms of this series are increasing, hence do not tend to zero, thus the series is divergent.
Now, perhaps your teacher meant $frac{n^n}{n!e^n}$, as then the series would still be divergent, but the terms do go to zero. However, there is a trick to make it work -- multiply the terms by $n$. By repeating a calculation like above, from $e<(1+1/n)^{n+1}$ it follows that the seqence $ncdotfrac{n^n}{n!e^n}$ is increasing, so all its terms are at least as large as the first one, equal to $1/e$. Hence $frac{n^n}{n!e^n}geqfrac{1}{en}$, and by divergence of harmonic series this series diverges too.
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
answered Jan 11 at 11:27
WojowuWojowu
17.8k22668
17.8k22668
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
add a comment |
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
Hey, thank you for the answer! Everything here seems working to me. During our classes we were using only versions with limits of all tests. I send your answer to my teacher and his response is that he has to think about it but he thinks there's a loophole here as what you just did is basicly a ratio test and it's inconclusive when the limit of the ratio as n approaches infinity equals 1. He says we have to prove divergence for all terms up to infinity and the limit to infinity here gives you 1 which is making the test inconclusive here.
$endgroup$
– Skyfall
Jan 12 at 9:49
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
$begingroup$
@Skyfall Thank you for sharing his response. What I am doing here is not so much a ratio test (though I see why it looks like it), but rather a simple comparison test, and ratios are used to decide the comparison. Focusing on just the first part, we have terms $a_1, a_2,dots$. I am taking ratios $a_{n+1}/a_n$, but my conclusion is not based on the fact that this ratio goes to $1$. Instead, we have $a_{n+1}/a_n>1$, from which I conclude $a_{n+1}>a_n>dots>a_1$. Hence $sum_{n=1}^ka_ngeqsum_{n=1}^ka_1=ka_1$ which clearly diverges. Hope this clarifies what's going on in the proof.
$endgroup$
– Wojowu
Jan 12 at 9:55
add a comment |
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It would be sufficient to show that the limit of the expression for $nrightarrow infty$ is not $0$, but I have no simple idea yet.
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– Peter
Jan 11 at 11:20
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Maybe you can use the Stirling's approximation
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– Ixion
Jan 11 at 11:27
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Hi and welcome to the M.SE. If you want to add beautifully looking and mathematical text to your questions/answers, you may want to have a look at the Math Jax quick reference: you already did a good job by providing context to your problem.
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– Daniele Tampieri
Jan 11 at 11:27
2
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@Ixion "he knows it diverges because of Stirling's formula but this is not quite our level yet."
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– Wojowu
Jan 11 at 11:28
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@Wojowu, ouch, thanks... My bad!
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– Ixion
Jan 11 at 11:29