What is the difference between convergence of a sequence and convergence of a series?
$begingroup$
I'm preparing for my calculus exam and I'm unsure how to approach the question: "Explain the difference between convergence of a sequence and convergence of a series?"
I understand the following:
Let the sequence $a_n$ exist such that $a_n =frac{1}{n^2}$
Then $lim_{ntoinfty} a_n=lim_{ntoinfty} frac{1}{n^2}=0$ therefore $a_n$ converges to $0$.
And the series $sum_{i=1}^{n}a_n=1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ... +frac{1}{n^2}$
And by the $n$-th term test, this series converges. But, I don't understand why or how the convergence between the series and the sequence is different.
I looked online and I find a lot of answers on how to determine convergence or divergence, but the only difference I've found is that you use limits to test sequences and series have more complex testing requirements. Please help!
calculus sequences-and-series convergence
$endgroup$
add a comment |
$begingroup$
I'm preparing for my calculus exam and I'm unsure how to approach the question: "Explain the difference between convergence of a sequence and convergence of a series?"
I understand the following:
Let the sequence $a_n$ exist such that $a_n =frac{1}{n^2}$
Then $lim_{ntoinfty} a_n=lim_{ntoinfty} frac{1}{n^2}=0$ therefore $a_n$ converges to $0$.
And the series $sum_{i=1}^{n}a_n=1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ... +frac{1}{n^2}$
And by the $n$-th term test, this series converges. But, I don't understand why or how the convergence between the series and the sequence is different.
I looked online and I find a lot of answers on how to determine convergence or divergence, but the only difference I've found is that you use limits to test sequences and series have more complex testing requirements. Please help!
calculus sequences-and-series convergence
$endgroup$
1
$begingroup$
This is kind of baffling to me. A series is just a sequence. Edit: I guess I should elaborate. A finite series is easily defined as simply "adding all the numbers" but an infinite series is not. So to my knowledge a series is usually defined as a sequence of its terms. And then the convergence/divergence of a series is the same as the convergence/divergence of its related sequence.
$endgroup$
– Zelzy
Nov 22 '14 at 23:33
add a comment |
$begingroup$
I'm preparing for my calculus exam and I'm unsure how to approach the question: "Explain the difference between convergence of a sequence and convergence of a series?"
I understand the following:
Let the sequence $a_n$ exist such that $a_n =frac{1}{n^2}$
Then $lim_{ntoinfty} a_n=lim_{ntoinfty} frac{1}{n^2}=0$ therefore $a_n$ converges to $0$.
And the series $sum_{i=1}^{n}a_n=1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ... +frac{1}{n^2}$
And by the $n$-th term test, this series converges. But, I don't understand why or how the convergence between the series and the sequence is different.
I looked online and I find a lot of answers on how to determine convergence or divergence, but the only difference I've found is that you use limits to test sequences and series have more complex testing requirements. Please help!
calculus sequences-and-series convergence
$endgroup$
I'm preparing for my calculus exam and I'm unsure how to approach the question: "Explain the difference between convergence of a sequence and convergence of a series?"
I understand the following:
Let the sequence $a_n$ exist such that $a_n =frac{1}{n^2}$
Then $lim_{ntoinfty} a_n=lim_{ntoinfty} frac{1}{n^2}=0$ therefore $a_n$ converges to $0$.
And the series $sum_{i=1}^{n}a_n=1 + frac{1}{4} + frac{1}{9} + frac{1}{16} + ... +frac{1}{n^2}$
And by the $n$-th term test, this series converges. But, I don't understand why or how the convergence between the series and the sequence is different.
I looked online and I find a lot of answers on how to determine convergence or divergence, but the only difference I've found is that you use limits to test sequences and series have more complex testing requirements. Please help!
calculus sequences-and-series convergence
calculus sequences-and-series convergence
edited Nov 22 '14 at 23:34
graydad
12.8k61934
12.8k61934
asked Nov 22 '14 at 23:28
Elliott PostElliott Post
2315
2315
1
$begingroup$
This is kind of baffling to me. A series is just a sequence. Edit: I guess I should elaborate. A finite series is easily defined as simply "adding all the numbers" but an infinite series is not. So to my knowledge a series is usually defined as a sequence of its terms. And then the convergence/divergence of a series is the same as the convergence/divergence of its related sequence.
$endgroup$
– Zelzy
Nov 22 '14 at 23:33
add a comment |
1
$begingroup$
This is kind of baffling to me. A series is just a sequence. Edit: I guess I should elaborate. A finite series is easily defined as simply "adding all the numbers" but an infinite series is not. So to my knowledge a series is usually defined as a sequence of its terms. And then the convergence/divergence of a series is the same as the convergence/divergence of its related sequence.
$endgroup$
– Zelzy
Nov 22 '14 at 23:33
1
1
$begingroup$
This is kind of baffling to me. A series is just a sequence. Edit: I guess I should elaborate. A finite series is easily defined as simply "adding all the numbers" but an infinite series is not. So to my knowledge a series is usually defined as a sequence of its terms. And then the convergence/divergence of a series is the same as the convergence/divergence of its related sequence.
$endgroup$
– Zelzy
Nov 22 '14 at 23:33
$begingroup$
This is kind of baffling to me. A series is just a sequence. Edit: I guess I should elaborate. A finite series is easily defined as simply "adding all the numbers" but an infinite series is not. So to my knowledge a series is usually defined as a sequence of its terms. And then the convergence/divergence of a series is the same as the convergence/divergence of its related sequence.
$endgroup$
– Zelzy
Nov 22 '14 at 23:33
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Series can be baffling things. The trouble is that the convergence of the terms tells you nothing about the convergence of the series. We know that $lim_{ntoinfty}a_n=0$ is necessary for a series $sum_{k=0}^infty a_n$ to converge, but it is not sufficient.
For example consider the sequences $a_n=1/n$, $b_n=(-1)^n/n$, $c_n=1/n^2$, $d_n=1/n^3$, $e_n=1/n^5$. All of these sequences converge to zero. But:
- $sum_{k=1}^infty a_n$ diverges (harmonic series)
- $sum_{k=1}^infty b_n$ converges (alternating series)
- $sum_{k=1}^infty c_n$ converges to $pi^2/6$
- $sum_{k=1}^infty d_n$ converges, to an irrational number (Apéry's theorem)
- $sum_{k=1}^infty e_n$ converges, but its not known if the result is rational or irrational.
There is, as far as I know, no complete decision method for convergence of a series. You try a sequence of tests, and each one will return a result of converge, diverge, undecided. If the latter, you try another test.
There are certain families of series, such as geometric series and $p$-series, for which convergence or divergence is trivial ($sum_{k=0}^infty r^k$ converges if and only if $|r|<1$), but in general, given a new series which doesn't fit into a known class, you're on your own.
As far as I know, this problem is unsolved: is there a sequence of rational numbers $a_n$ for which $lim_{ntoinfty}a_{n+1}/a_n=0$ and $sum_{n=0}^infty a_n=pi$? (Note that if we replace $pi$ by $e$ then the result is trivial, as we can put $a_n=1/n!$.)
As you see, series are tricky.
$endgroup$
add a comment |
$begingroup$
You can identify a series with the sequence of its partial sums:
$$
S_n = sum_{k=1}^n a_k.
$$
So everything you know about sequences can be applied to series, and vice-versa.
However dealing with series is usually more difficult because, in general, it can be very difficult to find the limit. This is due to the indirect definition of partial sums... So to deal with series and to prove their convergence, one should use methods which do not require the limit (i.e. the sum) of the series to be known in advance.
$endgroup$
add a comment |
$begingroup$
If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.
Convergence of a series $sum_{n=1}^infty a_n$ is simply the convergence of the sequence of partial sums $S_N = sum_{n=1}^N a_n$.
Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + sum_{n=1}^infty (a_{n+1} - a_n)$; note the $N$-th partial sum is
$$
a_1 + sum_{n=1}^N (a_{n+1} - a_n) = a_1 + sum_{n=2}^{N+1} a_n - sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N.
$$
However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.
$endgroup$
add a comment |
$begingroup$
You could say that all series are sequences, meaning that you could make a sequence taking its partial sums and checking whether or not it approaches a limit, but certainly not all sequences are series.
$endgroup$
add a comment |
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4 Answers
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4 Answers
4
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active
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$begingroup$
Series can be baffling things. The trouble is that the convergence of the terms tells you nothing about the convergence of the series. We know that $lim_{ntoinfty}a_n=0$ is necessary for a series $sum_{k=0}^infty a_n$ to converge, but it is not sufficient.
For example consider the sequences $a_n=1/n$, $b_n=(-1)^n/n$, $c_n=1/n^2$, $d_n=1/n^3$, $e_n=1/n^5$. All of these sequences converge to zero. But:
- $sum_{k=1}^infty a_n$ diverges (harmonic series)
- $sum_{k=1}^infty b_n$ converges (alternating series)
- $sum_{k=1}^infty c_n$ converges to $pi^2/6$
- $sum_{k=1}^infty d_n$ converges, to an irrational number (Apéry's theorem)
- $sum_{k=1}^infty e_n$ converges, but its not known if the result is rational or irrational.
There is, as far as I know, no complete decision method for convergence of a series. You try a sequence of tests, and each one will return a result of converge, diverge, undecided. If the latter, you try another test.
There are certain families of series, such as geometric series and $p$-series, for which convergence or divergence is trivial ($sum_{k=0}^infty r^k$ converges if and only if $|r|<1$), but in general, given a new series which doesn't fit into a known class, you're on your own.
As far as I know, this problem is unsolved: is there a sequence of rational numbers $a_n$ for which $lim_{ntoinfty}a_{n+1}/a_n=0$ and $sum_{n=0}^infty a_n=pi$? (Note that if we replace $pi$ by $e$ then the result is trivial, as we can put $a_n=1/n!$.)
As you see, series are tricky.
$endgroup$
add a comment |
$begingroup$
Series can be baffling things. The trouble is that the convergence of the terms tells you nothing about the convergence of the series. We know that $lim_{ntoinfty}a_n=0$ is necessary for a series $sum_{k=0}^infty a_n$ to converge, but it is not sufficient.
For example consider the sequences $a_n=1/n$, $b_n=(-1)^n/n$, $c_n=1/n^2$, $d_n=1/n^3$, $e_n=1/n^5$. All of these sequences converge to zero. But:
- $sum_{k=1}^infty a_n$ diverges (harmonic series)
- $sum_{k=1}^infty b_n$ converges (alternating series)
- $sum_{k=1}^infty c_n$ converges to $pi^2/6$
- $sum_{k=1}^infty d_n$ converges, to an irrational number (Apéry's theorem)
- $sum_{k=1}^infty e_n$ converges, but its not known if the result is rational or irrational.
There is, as far as I know, no complete decision method for convergence of a series. You try a sequence of tests, and each one will return a result of converge, diverge, undecided. If the latter, you try another test.
There are certain families of series, such as geometric series and $p$-series, for which convergence or divergence is trivial ($sum_{k=0}^infty r^k$ converges if and only if $|r|<1$), but in general, given a new series which doesn't fit into a known class, you're on your own.
As far as I know, this problem is unsolved: is there a sequence of rational numbers $a_n$ for which $lim_{ntoinfty}a_{n+1}/a_n=0$ and $sum_{n=0}^infty a_n=pi$? (Note that if we replace $pi$ by $e$ then the result is trivial, as we can put $a_n=1/n!$.)
As you see, series are tricky.
$endgroup$
add a comment |
$begingroup$
Series can be baffling things. The trouble is that the convergence of the terms tells you nothing about the convergence of the series. We know that $lim_{ntoinfty}a_n=0$ is necessary for a series $sum_{k=0}^infty a_n$ to converge, but it is not sufficient.
For example consider the sequences $a_n=1/n$, $b_n=(-1)^n/n$, $c_n=1/n^2$, $d_n=1/n^3$, $e_n=1/n^5$. All of these sequences converge to zero. But:
- $sum_{k=1}^infty a_n$ diverges (harmonic series)
- $sum_{k=1}^infty b_n$ converges (alternating series)
- $sum_{k=1}^infty c_n$ converges to $pi^2/6$
- $sum_{k=1}^infty d_n$ converges, to an irrational number (Apéry's theorem)
- $sum_{k=1}^infty e_n$ converges, but its not known if the result is rational or irrational.
There is, as far as I know, no complete decision method for convergence of a series. You try a sequence of tests, and each one will return a result of converge, diverge, undecided. If the latter, you try another test.
There are certain families of series, such as geometric series and $p$-series, for which convergence or divergence is trivial ($sum_{k=0}^infty r^k$ converges if and only if $|r|<1$), but in general, given a new series which doesn't fit into a known class, you're on your own.
As far as I know, this problem is unsolved: is there a sequence of rational numbers $a_n$ for which $lim_{ntoinfty}a_{n+1}/a_n=0$ and $sum_{n=0}^infty a_n=pi$? (Note that if we replace $pi$ by $e$ then the result is trivial, as we can put $a_n=1/n!$.)
As you see, series are tricky.
$endgroup$
Series can be baffling things. The trouble is that the convergence of the terms tells you nothing about the convergence of the series. We know that $lim_{ntoinfty}a_n=0$ is necessary for a series $sum_{k=0}^infty a_n$ to converge, but it is not sufficient.
For example consider the sequences $a_n=1/n$, $b_n=(-1)^n/n$, $c_n=1/n^2$, $d_n=1/n^3$, $e_n=1/n^5$. All of these sequences converge to zero. But:
- $sum_{k=1}^infty a_n$ diverges (harmonic series)
- $sum_{k=1}^infty b_n$ converges (alternating series)
- $sum_{k=1}^infty c_n$ converges to $pi^2/6$
- $sum_{k=1}^infty d_n$ converges, to an irrational number (Apéry's theorem)
- $sum_{k=1}^infty e_n$ converges, but its not known if the result is rational or irrational.
There is, as far as I know, no complete decision method for convergence of a series. You try a sequence of tests, and each one will return a result of converge, diverge, undecided. If the latter, you try another test.
There are certain families of series, such as geometric series and $p$-series, for which convergence or divergence is trivial ($sum_{k=0}^infty r^k$ converges if and only if $|r|<1$), but in general, given a new series which doesn't fit into a known class, you're on your own.
As far as I know, this problem is unsolved: is there a sequence of rational numbers $a_n$ for which $lim_{ntoinfty}a_{n+1}/a_n=0$ and $sum_{n=0}^infty a_n=pi$? (Note that if we replace $pi$ by $e$ then the result is trivial, as we can put $a_n=1/n!$.)
As you see, series are tricky.
answered Nov 23 '14 at 0:10
AlasdairAlasdair
418312
418312
add a comment |
add a comment |
$begingroup$
You can identify a series with the sequence of its partial sums:
$$
S_n = sum_{k=1}^n a_k.
$$
So everything you know about sequences can be applied to series, and vice-versa.
However dealing with series is usually more difficult because, in general, it can be very difficult to find the limit. This is due to the indirect definition of partial sums... So to deal with series and to prove their convergence, one should use methods which do not require the limit (i.e. the sum) of the series to be known in advance.
$endgroup$
add a comment |
$begingroup$
You can identify a series with the sequence of its partial sums:
$$
S_n = sum_{k=1}^n a_k.
$$
So everything you know about sequences can be applied to series, and vice-versa.
However dealing with series is usually more difficult because, in general, it can be very difficult to find the limit. This is due to the indirect definition of partial sums... So to deal with series and to prove their convergence, one should use methods which do not require the limit (i.e. the sum) of the series to be known in advance.
$endgroup$
add a comment |
$begingroup$
You can identify a series with the sequence of its partial sums:
$$
S_n = sum_{k=1}^n a_k.
$$
So everything you know about sequences can be applied to series, and vice-versa.
However dealing with series is usually more difficult because, in general, it can be very difficult to find the limit. This is due to the indirect definition of partial sums... So to deal with series and to prove their convergence, one should use methods which do not require the limit (i.e. the sum) of the series to be known in advance.
$endgroup$
You can identify a series with the sequence of its partial sums:
$$
S_n = sum_{k=1}^n a_k.
$$
So everything you know about sequences can be applied to series, and vice-versa.
However dealing with series is usually more difficult because, in general, it can be very difficult to find the limit. This is due to the indirect definition of partial sums... So to deal with series and to prove their convergence, one should use methods which do not require the limit (i.e. the sum) of the series to be known in advance.
answered Nov 22 '14 at 23:39
Emanuele PaoliniEmanuele Paolini
17.9k22052
17.9k22052
add a comment |
add a comment |
$begingroup$
If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.
Convergence of a series $sum_{n=1}^infty a_n$ is simply the convergence of the sequence of partial sums $S_N = sum_{n=1}^N a_n$.
Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + sum_{n=1}^infty (a_{n+1} - a_n)$; note the $N$-th partial sum is
$$
a_1 + sum_{n=1}^N (a_{n+1} - a_n) = a_1 + sum_{n=2}^{N+1} a_n - sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N.
$$
However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.
$endgroup$
add a comment |
$begingroup$
If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.
Convergence of a series $sum_{n=1}^infty a_n$ is simply the convergence of the sequence of partial sums $S_N = sum_{n=1}^N a_n$.
Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + sum_{n=1}^infty (a_{n+1} - a_n)$; note the $N$-th partial sum is
$$
a_1 + sum_{n=1}^N (a_{n+1} - a_n) = a_1 + sum_{n=2}^{N+1} a_n - sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N.
$$
However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.
$endgroup$
add a comment |
$begingroup$
If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.
Convergence of a series $sum_{n=1}^infty a_n$ is simply the convergence of the sequence of partial sums $S_N = sum_{n=1}^N a_n$.
Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + sum_{n=1}^infty (a_{n+1} - a_n)$; note the $N$-th partial sum is
$$
a_1 + sum_{n=1}^N (a_{n+1} - a_n) = a_1 + sum_{n=2}^{N+1} a_n - sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N.
$$
However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.
$endgroup$
If we are talking about sequences and series of real or complex numbers, or of vectors in a real (or complex) normed vector space, then convergence of sequences and series are equivalent concepts.
Convergence of a series $sum_{n=1}^infty a_n$ is simply the convergence of the sequence of partial sums $S_N = sum_{n=1}^N a_n$.
Convergence of a sequence $(a_n)$ of numbers is equivalent to the convergence of the series $a_1 + sum_{n=1}^infty (a_{n+1} - a_n)$; note the $N$-th partial sum is
$$
a_1 + sum_{n=1}^N (a_{n+1} - a_n) = a_1 + sum_{n=2}^{N+1} a_n - sum_{n=1}^N a_n = a_1 + a_N - a_1 = a_N.
$$
However, in the more general setting of topology the notion of a sequence is more general than the notion of a series, because in topological spaces without a vector space structure the notion of a series makes no sense (how will you define a series if you can't add things?). Therefore it may be convenient to think of sequences as the more natural idea, and series as a special case of sequences.
answered Nov 23 '14 at 0:18
Gyu Eun LeeGyu Eun Lee
13.4k2353
13.4k2353
add a comment |
add a comment |
$begingroup$
You could say that all series are sequences, meaning that you could make a sequence taking its partial sums and checking whether or not it approaches a limit, but certainly not all sequences are series.
$endgroup$
add a comment |
$begingroup$
You could say that all series are sequences, meaning that you could make a sequence taking its partial sums and checking whether or not it approaches a limit, but certainly not all sequences are series.
$endgroup$
add a comment |
$begingroup$
You could say that all series are sequences, meaning that you could make a sequence taking its partial sums and checking whether or not it approaches a limit, but certainly not all sequences are series.
$endgroup$
You could say that all series are sequences, meaning that you could make a sequence taking its partial sums and checking whether or not it approaches a limit, but certainly not all sequences are series.
answered Jan 17 at 23:28
Juan123Juan123
948
948
add a comment |
add a comment |
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1
$begingroup$
This is kind of baffling to me. A series is just a sequence. Edit: I guess I should elaborate. A finite series is easily defined as simply "adding all the numbers" but an infinite series is not. So to my knowledge a series is usually defined as a sequence of its terms. And then the convergence/divergence of a series is the same as the convergence/divergence of its related sequence.
$endgroup$
– Zelzy
Nov 22 '14 at 23:33