Does $sum _{n=1}^{infty ::}left(frac{-3}{4}right)^{n+1}$ converge?
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This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?
sequences-and-series limits
edited Nov 23 at 3:28
Seth
43612
asked Nov 23 at 3:23
Fourth
464
add a comment |
up vote
2
down vote
favorite
This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?
sequences-and-series limits
edited Nov 23 at 3:28
Seth
43612
asked Nov 23 at 3:23
Fourth
464
4
It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46
Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?
sequences-and-series limits
edited Nov 23 at 3:28
Seth
43612
asked Nov 23 at 3:23
Fourth
464
This is a geometric series, where $a=(3/4)^2$ and $r=-3/4$, meaning that it converges to $9/28$ right? The answer in my textbook says that it diverges, whereas Symbolab says it converges absolutely. So does it converge or diverge?
sequences-and-series limits
sequences-and-series limits
edited Nov 23 at 3:28
Seth
43612
asked Nov 23 at 3:23
Fourth
464
edited Nov 23 at 3:28
Seth
43612
asked Nov 23 at 3:23
Fourth
464
edited Nov 23 at 3:28
Seth
43612
edited Nov 23 at 3:28
Seth
43612
edited Nov 23 at 3:28
Seth
43612
43612
asked Nov 23 at 3:23
Fourth
464
asked Nov 23 at 3:23
Fourth
464
asked Nov 23 at 3:23
Fourth
464
464
4
It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46
Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25
add a comment |
4
It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46
Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25
4
4
It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26
It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46
Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25
Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25
add a comment |
4 Answers
4
active
oldest
votes
up vote
2
down vote
This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
answered Nov 23 at 3:29
Juan123
815
add a comment |
up vote
1
down vote
If $|x| < 1$ then
$displaystylesum_{n=0}^{infty} x^n
=frac1{1-x}$.
Therefore
$begin{align}\
sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
&=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
&=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
&=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
&=frac{9}{16}frac{4}{4+3}\
&=frac{9}{4cdot 7}\
&=frac{9}{28}\
end{align}
$
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 4:14
marty cohen
72k547126
add a comment |
up vote
1
down vote
The series can be written as the sum of positive terms and negative terms.
$$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 3:52
Ajit Sharma
213
add a comment |
up vote
1
down vote
The Leibniz criterion could be mentioned.
The series is convergent.
https://en.m.wikipedia.org/wiki/Alternating_series_test
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
answered Nov 23 at 3:29
Juan123
815
add a comment |
up vote
2
down vote
This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
answered Nov 23 at 3:29
Juan123
815
add a comment |
up vote
2
down vote
up vote
2
down vote
This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
answered Nov 23 at 3:29
Juan123
815
This is an absolutely convergent series, therefore the series has to converge regardless of the minus signs.
answered Nov 23 at 3:29
Juan123
815
answered Nov 23 at 3:29
Juan123
815
answered Nov 23 at 3:29
Juan123
815
answered Nov 23 at 3:29
Juan123
815
815
add a comment |
add a comment |
up vote
1
down vote
If $|x| < 1$ then
$displaystylesum_{n=0}^{infty} x^n
=frac1{1-x}$.
Therefore
$begin{align}\
sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
&=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
&=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
&=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
&=frac{9}{16}frac{4}{4+3}\
&=frac{9}{4cdot 7}\
&=frac{9}{28}\
end{align}
$
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 4:14
marty cohen
72k547126
add a comment |
up vote
1
down vote
If $|x| < 1$ then
$displaystylesum_{n=0}^{infty} x^n
=frac1{1-x}$.
Therefore
$begin{align}\
sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
&=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
&=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
&=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
&=frac{9}{16}frac{4}{4+3}\
&=frac{9}{4cdot 7}\
&=frac{9}{28}\
end{align}
$
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 4:14
marty cohen
72k547126
add a comment |
up vote
1
down vote
up vote
1
down vote
If $|x| < 1$ then
$displaystylesum_{n=0}^{infty} x^n
=frac1{1-x}$.
Therefore
$begin{align}\
sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
&=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
&=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
&=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
&=frac{9}{16}frac{4}{4+3}\
&=frac{9}{4cdot 7}\
&=frac{9}{28}\
end{align}
$
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 4:14
marty cohen
72k547126
If $|x| < 1$ then
$displaystylesum_{n=0}^{infty} x^n
=frac1{1-x}$.
Therefore
$begin{align}\
sum_{n=1}^{infty} left(frac{-3}{4}right)^{n+1}
&=left(frac{-3}{4}right)^2sum_{n=1}^{infty} left(frac{-3}{4}right)^{n-1}\
&=frac{9}{16}sum_{n=0}^{infty} left(frac{-3}{4}right)^{n}\
&=frac{9}{16}frac1{1-left(frac{-3}{4}right)}\
&=frac{9}{16}frac{4}{4+3}\
&=frac{9}{4cdot 7}\
&=frac{9}{28}\
end{align}
$
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 4:14
marty cohen
72k547126
edited Nov 23 at 4:16
Chickenmancer
3,299723
edited Nov 23 at 4:16
Chickenmancer
3,299723
edited Nov 23 at 4:16
Chickenmancer
3,299723
3,299723
answered Nov 23 at 4:14
marty cohen
72k547126
answered Nov 23 at 4:14
marty cohen
72k547126
answered Nov 23 at 4:14
marty cohen
72k547126
72k547126
add a comment |
add a comment |
up vote
1
down vote
The series can be written as the sum of positive terms and negative terms.
$$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 3:52
Ajit Sharma
213
add a comment |
up vote
1
down vote
The series can be written as the sum of positive terms and negative terms.
$$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 3:52
Ajit Sharma
213
add a comment |
up vote
1
down vote
up vote
1
down vote
The series can be written as the sum of positive terms and negative terms.
$$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 3:52
Ajit Sharma
213
The series can be written as the sum of positive terms and negative terms.
$$sum_{i=1}^{infty}(-3/4)^{i+1} = sum_{k=0}^{infty}(3/4)^{2k}-sum_{k=0}^{infty}(3/4)^{2k+1} = a_n - b_n$$
Both $a_n$ and $b_n$ being geometric series are convergent.
Hence, their sum should also be a convergent series. Proof of convergence of sum of two convergent series.
edited Nov 23 at 4:16
Chickenmancer
3,299723
answered Nov 23 at 3:52
Ajit Sharma
213
edited Nov 23 at 4:16
Chickenmancer
3,299723
edited Nov 23 at 4:16
Chickenmancer
3,299723
edited Nov 23 at 4:16
Chickenmancer
3,299723
3,299723
answered Nov 23 at 3:52
Ajit Sharma
213
answered Nov 23 at 3:52
Ajit Sharma
213
answered Nov 23 at 3:52
Ajit Sharma
213
213
add a comment |
add a comment |
up vote
1
down vote
The Leibniz criterion could be mentioned.
The series is convergent.
https://en.m.wikipedia.org/wiki/Alternating_series_test
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
add a comment |
up vote
1
down vote
The Leibniz criterion could be mentioned.
The series is convergent.
https://en.m.wikipedia.org/wiki/Alternating_series_test
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
add a comment |
up vote
1
down vote
up vote
1
down vote
The Leibniz criterion could be mentioned.
The series is convergent.
https://en.m.wikipedia.org/wiki/Alternating_series_test
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
The Leibniz criterion could be mentioned.
The series is convergent.
https://en.m.wikipedia.org/wiki/Alternating_series_test
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
answered Nov 23 at 9:01
Peter Szilas
10.6k2720
10.6k2720
add a comment |
add a comment |
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4
It most definitely converges since it's a geometric series whose ratio has absolute value less than $1$.
– Fimpellizieri
Nov 23 at 3:26
Absolute convergence is a very strong result in analysis. So basically, IF you can show that $sum_{n=1}^infty |a_n|$ converges, then you know the sum without absolute value $sum_{n=1}^infty a_n$ WILL converge ALSO. It makes showing convergence of some sums much easier.
– Decaf-Math
Nov 23 at 3:46
Your textbook probably has a typo.
– Paramanand Singh
Nov 23 at 12:25