Limit of a series exists but series diverges? The 2019 Stack Overflow Developer Survey Results Are InWhat is the limit for this seriesCan the alternating series test tell me that a series diverges?Testing A Series For ConvergenceInfinite series convergence testProving a series divergesProve wether or not the following series diverges or converges: $sum_n=0^infty (-1)^nnover n+1$Infinite Series Sum: $sum_n=1^infty frac 12^ntanleft(fracx2^nright) = frac1x - cot x$Show using the comparison test whether the following series divergesDetermine whether the series $frace^frac1nn^2$ converges or divergesDetermining if a limit of a infinite series exists
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Limit of a series exists but series diverges?
The 2019 Stack Overflow Developer Survey Results Are InWhat is the limit for this seriesCan the alternating series test tell me that a series diverges?Testing A Series For ConvergenceInfinite series convergence testProving a series divergesProve wether or not the following series diverges or converges: $sum_n=0^infty (-1)^nnover n+1$Infinite Series Sum: $sum_n=1^infty frac 12^ntanleft(fracx2^nright) = frac1x - cot x$Show using the comparison test whether the following series divergesDetermine whether the series $frace^frac1nn^2$ converges or divergesDetermining if a limit of a infinite series exists
$begingroup$
In finding the sum of the series
$$sum_n=1^infty frac4(4n-3)(4n+1)$$
I get to $$1-frac14n+1$$
and I just take $$lim_ntoinfty a_n = 1$$
which tells me that the sum of the series as n goes to infinity is 1. However, there's a theorem in the book I'm using (Thomas Calculus) that states that
$sum_n=1^infty a_n$ diverges if $lim_ntoinfty a_n$ fails to exist or is different from zero. T
So this would lead me to believe that the sum of the series is 1 but that it diverges. This doesn't make sense to me. Any help would be greatly appreciated!
EDIT: I see what my problem was now. I was writing out the first few terms in order to do telescoping to get the sum. However, the limit that I was taking was of the sum, NOT of the nth term. The nth term should go to zero. This makes sense. Thank you all for clarifying
sequences-and-series limits convergence
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
asked Mar 31 '18 at 19:08
НектоНекто
618
$endgroup$
add a comment |
$begingroup$
In finding the sum of the series
$$sum_n=1^infty frac4(4n-3)(4n+1)$$
I get to $$1-frac14n+1$$
and I just take $$lim_ntoinfty a_n = 1$$
which tells me that the sum of the series as n goes to infinity is 1. However, there's a theorem in the book I'm using (Thomas Calculus) that states that
$sum_n=1^infty a_n$ diverges if $lim_ntoinfty a_n$ fails to exist or is different from zero. T
So this would lead me to believe that the sum of the series is 1 but that it diverges. This doesn't make sense to me. Any help would be greatly appreciated!
EDIT: I see what my problem was now. I was writing out the first few terms in order to do telescoping to get the sum. However, the limit that I was taking was of the sum, NOT of the nth term. The nth term should go to zero. This makes sense. Thank you all for clarifying
sequences-and-series limits convergence
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
asked Mar 31 '18 at 19:08
НектоНекто
618
$endgroup$
add a comment |
$begingroup$
In finding the sum of the series
$$sum_n=1^infty frac4(4n-3)(4n+1)$$
I get to $$1-frac14n+1$$
and I just take $$lim_ntoinfty a_n = 1$$
which tells me that the sum of the series as n goes to infinity is 1. However, there's a theorem in the book I'm using (Thomas Calculus) that states that
$sum_n=1^infty a_n$ diverges if $lim_ntoinfty a_n$ fails to exist or is different from zero. T
So this would lead me to believe that the sum of the series is 1 but that it diverges. This doesn't make sense to me. Any help would be greatly appreciated!
EDIT: I see what my problem was now. I was writing out the first few terms in order to do telescoping to get the sum. However, the limit that I was taking was of the sum, NOT of the nth term. The nth term should go to zero. This makes sense. Thank you all for clarifying
sequences-and-series limits convergence
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
asked Mar 31 '18 at 19:08
НектоНекто
618
$endgroup$
In finding the sum of the series
$$sum_n=1^infty frac4(4n-3)(4n+1)$$
I get to $$1-frac14n+1$$
and I just take $$lim_ntoinfty a_n = 1$$
which tells me that the sum of the series as n goes to infinity is 1. However, there's a theorem in the book I'm using (Thomas Calculus) that states that
$sum_n=1^infty a_n$ diverges if $lim_ntoinfty a_n$ fails to exist or is different from zero. T
So this would lead me to believe that the sum of the series is 1 but that it diverges. This doesn't make sense to me. Any help would be greatly appreciated!
EDIT: I see what my problem was now. I was writing out the first few terms in order to do telescoping to get the sum. However, the limit that I was taking was of the sum, NOT of the nth term. The nth term should go to zero. This makes sense. Thank you all for clarifying
sequences-and-series limits convergence
sequences-and-series limits convergence
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
asked Mar 31 '18 at 19:08
НектоНекто
618
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
asked Mar 31 '18 at 19:08
НектоНекто
618
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
edited Mar 22 at 22:01
José Carlos Santos
173k23133242
173k23133242
asked Mar 31 '18 at 19:08
НектоНекто
618
asked Mar 31 '18 at 19:08
НектоНекто
618
asked Mar 31 '18 at 19:08
НектоНекто
618
618
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your theorem involves the limit of the terms, not the limit of the sum. Here $lim_n to infty frac 1(4n-3)(4n+1)=0$ so you have no problem. This does not guarantee that the sum converges, but it allows it.
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The series converges. I think that the theorem that you have in mind is about the limit $lim_ntoinftyfrac1(4n-3)(4n+1)$, which is $0$. Therefore, there is no contradiction.
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
add a comment |
$begingroup$
Note that we represent $a_n$ as $n$-th term of the sequence. So, the correct equality is $$lim_n to infty a_n=lim_n to infty frac 4(4n-1)(4n+3)=0$$
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your theorem involves the limit of the terms, not the limit of the sum. Here $lim_n to infty frac 1(4n-3)(4n+1)=0$ so you have no problem. This does not guarantee that the sum converges, but it allows it.
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Your theorem involves the limit of the terms, not the limit of the sum. Here $lim_n to infty frac 1(4n-3)(4n+1)=0$ so you have no problem. This does not guarantee that the sum converges, but it allows it.
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Your theorem involves the limit of the terms, not the limit of the sum. Here $lim_n to infty frac 1(4n-3)(4n+1)=0$ so you have no problem. This does not guarantee that the sum converges, but it allows it.
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Your theorem involves the limit of the terms, not the limit of the sum. Here $lim_n to infty frac 1(4n-3)(4n+1)=0$ so you have no problem. This does not guarantee that the sum converges, but it allows it.
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
answered Mar 31 '18 at 19:11
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
$begingroup$
The series converges. I think that the theorem that you have in mind is about the limit $lim_ntoinftyfrac1(4n-3)(4n+1)$, which is $0$. Therefore, there is no contradiction.
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
add a comment |
$begingroup$
The series converges. I think that the theorem that you have in mind is about the limit $lim_ntoinftyfrac1(4n-3)(4n+1)$, which is $0$. Therefore, there is no contradiction.
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
add a comment |
$begingroup$
The series converges. I think that the theorem that you have in mind is about the limit $lim_ntoinftyfrac1(4n-3)(4n+1)$, which is $0$. Therefore, there is no contradiction.
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
$endgroup$
The series converges. I think that the theorem that you have in mind is about the limit $lim_ntoinftyfrac1(4n-3)(4n+1)$, which is $0$. Therefore, there is no contradiction.
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
answered Mar 31 '18 at 19:11
José Carlos SantosJosé Carlos Santos
173k23133242
173k23133242
add a comment |
add a comment |
$begingroup$
Note that we represent $a_n$ as $n$-th term of the sequence. So, the correct equality is $$lim_n to infty a_n=lim_n to infty frac 4(4n-1)(4n+3)=0$$
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
$endgroup$
add a comment |
$begingroup$
Note that we represent $a_n$ as $n$-th term of the sequence. So, the correct equality is $$lim_n to infty a_n=lim_n to infty frac 4(4n-1)(4n+3)=0$$
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
$endgroup$
add a comment |
$begingroup$
Note that we represent $a_n$ as $n$-th term of the sequence. So, the correct equality is $$lim_n to infty a_n=lim_n to infty frac 4(4n-1)(4n+3)=0$$
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
$endgroup$
Note that we represent $a_n$ as $n$-th term of the sequence. So, the correct equality is $$lim_n to infty a_n=lim_n to infty frac 4(4n-1)(4n+3)=0$$
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
answered Mar 31 '18 at 19:11
Jaideep KhareJaideep Khare
17.8k32669
17.8k32669
add a comment |
add a comment |
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