Mixing
Cesaro Mean of Sequences - Convergence
Clash Royale CLAN TAG#URR8PPP
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Show that if $(x_n)$ is a convergent sequence then the sequence given by the averages $$y_n = fracx_1 + x_2 + cdots + x_nn$$ also converges to the same limit.
Attempt at Proof.
Since $(x_n)$ converges we can say that for all $m$, such that $m ge N Rightarrow |x_n - L|lt epsilon$.
Base Case. Let n=1 and we have for all m such that $m ge N_0 Rightarrow |x_1 - L|lt epsilon$ and also for $m ge N_1 Rightarrow |x_n+1 - L|lt epsilon$.
Induction Hypothesis. Assume that for an appropriate choice of $N_2$ we have for all m, $m ge N_2 Rightarrow |y_n - L|lt epsilon$.
Choose $maxN_1,N_2$ such that for all $m ge maxN_1,N_2$ implies $$|fracx_1 + x_2 + cdots + x_nn- L|+ |x_n+1 - L|lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n-2L |lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L |le |fracx_1 + x_2 + cdots + x_n+ n x_n+12n-L |lt epsilon$$
and
$$|fracx_1 + x_2 + cdots + x_n+ x_n+1n+1-L |lt|fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L| lt epsilon.$$
Is this approach correct? If not, can you please provide the correct proof. I feel as though I made a mistake in the calculations. Thanks in advance.
real-analysis sequences-and-series proof-verification induction
asked Jul 21 at 23:13
Red
1,747733
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up vote
1
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Show that if $(x_n)$ is a convergent sequence then the sequence given by the averages $$y_n = fracx_1 + x_2 + cdots + x_nn$$ also converges to the same limit.
Attempt at Proof.
Since $(x_n)$ converges we can say that for all $m$, such that $m ge N Rightarrow |x_n - L|lt epsilon$.
Base Case. Let n=1 and we have for all m such that $m ge N_0 Rightarrow |x_1 - L|lt epsilon$ and also for $m ge N_1 Rightarrow |x_n+1 - L|lt epsilon$.
Induction Hypothesis. Assume that for an appropriate choice of $N_2$ we have for all m, $m ge N_2 Rightarrow |y_n - L|lt epsilon$.
Choose $maxN_1,N_2$ such that for all $m ge maxN_1,N_2$ implies $$|fracx_1 + x_2 + cdots + x_nn- L|+ |x_n+1 - L|lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n-2L |lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L |le |fracx_1 + x_2 + cdots + x_n+ n x_n+12n-L |lt epsilon$$
and
$$|fracx_1 + x_2 + cdots + x_n+ x_n+1n+1-L |lt|fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L| lt epsilon.$$
Is this approach correct? If not, can you please provide the correct proof. I feel as though I made a mistake in the calculations. Thanks in advance.
real-analysis sequences-and-series proof-verification induction
asked Jul 21 at 23:13
Red
1,747733
see also this related post
– G Cab
Jul 22 at 0:31
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that if $(x_n)$ is a convergent sequence then the sequence given by the averages $$y_n = fracx_1 + x_2 + cdots + x_nn$$ also converges to the same limit.
Attempt at Proof.
Since $(x_n)$ converges we can say that for all $m$, such that $m ge N Rightarrow |x_n - L|lt epsilon$.
Base Case. Let n=1 and we have for all m such that $m ge N_0 Rightarrow |x_1 - L|lt epsilon$ and also for $m ge N_1 Rightarrow |x_n+1 - L|lt epsilon$.
Induction Hypothesis. Assume that for an appropriate choice of $N_2$ we have for all m, $m ge N_2 Rightarrow |y_n - L|lt epsilon$.
Choose $maxN_1,N_2$ such that for all $m ge maxN_1,N_2$ implies $$|fracx_1 + x_2 + cdots + x_nn- L|+ |x_n+1 - L|lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n-2L |lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L |le |fracx_1 + x_2 + cdots + x_n+ n x_n+12n-L |lt epsilon$$
and
$$|fracx_1 + x_2 + cdots + x_n+ x_n+1n+1-L |lt|fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L| lt epsilon.$$
Is this approach correct? If not, can you please provide the correct proof. I feel as though I made a mistake in the calculations. Thanks in advance.
real-analysis sequences-and-series proof-verification induction
asked Jul 21 at 23:13
Red
1,747733
Show that if $(x_n)$ is a convergent sequence then the sequence given by the averages $$y_n = fracx_1 + x_2 + cdots + x_nn$$ also converges to the same limit.
Attempt at Proof.
Since $(x_n)$ converges we can say that for all $m$, such that $m ge N Rightarrow |x_n - L|lt epsilon$.
Base Case. Let n=1 and we have for all m such that $m ge N_0 Rightarrow |x_1 - L|lt epsilon$ and also for $m ge N_1 Rightarrow |x_n+1 - L|lt epsilon$.
Induction Hypothesis. Assume that for an appropriate choice of $N_2$ we have for all m, $m ge N_2 Rightarrow |y_n - L|lt epsilon$.
Choose $maxN_1,N_2$ such that for all $m ge maxN_1,N_2$ implies $$|fracx_1 + x_2 + cdots + x_nn- L|+ |x_n+1 - L|lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n-2L |lt 2epsilon$$
$$= |fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L |le |fracx_1 + x_2 + cdots + x_n+ n x_n+12n-L |lt epsilon$$
and
$$|fracx_1 + x_2 + cdots + x_n+ x_n+1n+1-L |lt|fracx_1 + x_2 + cdots + x_n+ n x_n+1n+1-L| lt epsilon.$$
Is this approach correct? If not, can you please provide the correct proof. I feel as though I made a mistake in the calculations. Thanks in advance.
real-analysis sequences-and-series proof-verification induction
asked Jul 21 at 23:13
Red
1,747733
asked Jul 21 at 23:13
Red
1,747733
asked Jul 21 at 23:13
Red
1,747733
asked Jul 21 at 23:13
Red
1,747733
1,747733
see also this related post
– G Cab
Jul 22 at 0:31
add a comment |Â
see also this related post
– G Cab
Jul 22 at 0:31
see also this related post
– G Cab
Jul 22 at 0:31
see also this related post
– G Cab
Jul 22 at 0:31
add a comment |Â
2 Answers
2
active
oldest
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up vote
3
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accepted
Your approach is wrong. Induction cannaot be used here unless you can get $N$ depending only on $epsilon $ and not on $n$. Here is a correct proof: $|y_n-L|=|frac x_1-L+x_2-L+...+X_n-L n|leq frac X_n-L n$ Split this into two sums: $frac +...+ n +frac x_k+1-L n$ Choose $k$ such that $|x_i-L|<epsilon $ for $i>k$. Then the second term is less than $frac epsilon +epsilon +... +epsilon n=frac n-k n epsilon <epsilon $. The first term tends to $0$ as $n to infty $ (because the numerator does not depend on $n$). We are done.
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
add a comment |Â
up vote
1
down vote
The intuition for a correct proof of this fact is as follows:
For any $epsilon>0$, there exists $N$ so that $lvert x_n-Lrvert<epsilon$ for all $ngeq N$. Equivalently, we can break the sequence into two parts:
Some initial segment $x_1,x_2,ldots, x_N-1$ of terms that can be anything (but there is only a fixed number of them), and
A tail $x_N+1,x_N+2,ldots$ of terms that are all close to (read: within $epsilon$ of) $L$.
If you pick some giant $n$, you get
$$
fracx_1+x_2+cdots+x_nn=fracx_1+cdots+x_Nn+fracx_N+1+x_N+2+cdots+x_nn.
$$
The first term has a fixed numerator, so it tends to $0$ as $ntoinfty$. (The average makes those few initial terms meaningless in the long run.) The second term can easily be seen to satisfy
$$
(L-epsilon)fracn-Nnleqfracx_N+1+x_N+2+cdots+x_nnleq(L+epsilon)fracn-Nn,
$$
and those bounds tend to $L-epsilon$ and $L+epsilon$ as $ntoinfty$. (Taking the average of a bunch of terms that are close to $L$ should yield a result close to $L$.)
Can you use these ingredients to complete a proof of the result? The intuition is all there.
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your approach is wrong. Induction cannaot be used here unless you can get $N$ depending only on $epsilon $ and not on $n$. Here is a correct proof: $|y_n-L|=|frac x_1-L+x_2-L+...+X_n-L n|leq frac X_n-L n$ Split this into two sums: $frac +...+ n +frac x_k+1-L n$ Choose $k$ such that $|x_i-L|<epsilon $ for $i>k$. Then the second term is less than $frac epsilon +epsilon +... +epsilon n=frac n-k n epsilon <epsilon $. The first term tends to $0$ as $n to infty $ (because the numerator does not depend on $n$). We are done.
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
add a comment |Â
up vote
3
down vote
accepted
Your approach is wrong. Induction cannaot be used here unless you can get $N$ depending only on $epsilon $ and not on $n$. Here is a correct proof: $|y_n-L|=|frac x_1-L+x_2-L+...+X_n-L n|leq frac X_n-L n$ Split this into two sums: $frac +...+ n +frac x_k+1-L n$ Choose $k$ such that $|x_i-L|<epsilon $ for $i>k$. Then the second term is less than $frac epsilon +epsilon +... +epsilon n=frac n-k n epsilon <epsilon $. The first term tends to $0$ as $n to infty $ (because the numerator does not depend on $n$). We are done.
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your approach is wrong. Induction cannaot be used here unless you can get $N$ depending only on $epsilon $ and not on $n$. Here is a correct proof: $|y_n-L|=|frac x_1-L+x_2-L+...+X_n-L n|leq frac X_n-L n$ Split this into two sums: $frac +...+ n +frac x_k+1-L n$ Choose $k$ such that $|x_i-L|<epsilon $ for $i>k$. Then the second term is less than $frac epsilon +epsilon +... +epsilon n=frac n-k n epsilon <epsilon $. The first term tends to $0$ as $n to infty $ (because the numerator does not depend on $n$). We are done.
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
Your approach is wrong. Induction cannaot be used here unless you can get $N$ depending only on $epsilon $ and not on $n$. Here is a correct proof: $|y_n-L|=|frac x_1-L+x_2-L+...+X_n-L n|leq frac X_n-L n$ Split this into two sums: $frac +...+ n +frac x_k+1-L n$ Choose $k$ such that $|x_i-L|<epsilon $ for $i>k$. Then the second term is less than $frac epsilon +epsilon +... +epsilon n=frac n-k n epsilon <epsilon $. The first term tends to $0$ as $n to infty $ (because the numerator does not depend on $n$). We are done.
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 23:27
Kavi Rama Murthy
20.6k2830
20.6k2830
add a comment |Â
add a comment |Â
up vote
1
down vote
The intuition for a correct proof of this fact is as follows:
For any $epsilon>0$, there exists $N$ so that $lvert x_n-Lrvert<epsilon$ for all $ngeq N$. Equivalently, we can break the sequence into two parts:
Some initial segment $x_1,x_2,ldots, x_N-1$ of terms that can be anything (but there is only a fixed number of them), and
A tail $x_N+1,x_N+2,ldots$ of terms that are all close to (read: within $epsilon$ of) $L$.
If you pick some giant $n$, you get
$$
fracx_1+x_2+cdots+x_nn=fracx_1+cdots+x_Nn+fracx_N+1+x_N+2+cdots+x_nn.
$$
The first term has a fixed numerator, so it tends to $0$ as $ntoinfty$. (The average makes those few initial terms meaningless in the long run.) The second term can easily be seen to satisfy
$$
(L-epsilon)fracn-Nnleqfracx_N+1+x_N+2+cdots+x_nnleq(L+epsilon)fracn-Nn,
$$
and those bounds tend to $L-epsilon$ and $L+epsilon$ as $ntoinfty$. (Taking the average of a bunch of terms that are close to $L$ should yield a result close to $L$.)
Can you use these ingredients to complete a proof of the result? The intuition is all there.
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
add a comment |Â
up vote
1
down vote
The intuition for a correct proof of this fact is as follows:
For any $epsilon>0$, there exists $N$ so that $lvert x_n-Lrvert<epsilon$ for all $ngeq N$. Equivalently, we can break the sequence into two parts:
Some initial segment $x_1,x_2,ldots, x_N-1$ of terms that can be anything (but there is only a fixed number of them), and
A tail $x_N+1,x_N+2,ldots$ of terms that are all close to (read: within $epsilon$ of) $L$.
If you pick some giant $n$, you get
$$
fracx_1+x_2+cdots+x_nn=fracx_1+cdots+x_Nn+fracx_N+1+x_N+2+cdots+x_nn.
$$
The first term has a fixed numerator, so it tends to $0$ as $ntoinfty$. (The average makes those few initial terms meaningless in the long run.) The second term can easily be seen to satisfy
$$
(L-epsilon)fracn-Nnleqfracx_N+1+x_N+2+cdots+x_nnleq(L+epsilon)fracn-Nn,
$$
and those bounds tend to $L-epsilon$ and $L+epsilon$ as $ntoinfty$. (Taking the average of a bunch of terms that are close to $L$ should yield a result close to $L$.)
Can you use these ingredients to complete a proof of the result? The intuition is all there.
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The intuition for a correct proof of this fact is as follows:
For any $epsilon>0$, there exists $N$ so that $lvert x_n-Lrvert<epsilon$ for all $ngeq N$. Equivalently, we can break the sequence into two parts:
Some initial segment $x_1,x_2,ldots, x_N-1$ of terms that can be anything (but there is only a fixed number of them), and
A tail $x_N+1,x_N+2,ldots$ of terms that are all close to (read: within $epsilon$ of) $L$.
If you pick some giant $n$, you get
$$
fracx_1+x_2+cdots+x_nn=fracx_1+cdots+x_Nn+fracx_N+1+x_N+2+cdots+x_nn.
$$
The first term has a fixed numerator, so it tends to $0$ as $ntoinfty$. (The average makes those few initial terms meaningless in the long run.) The second term can easily be seen to satisfy
$$
(L-epsilon)fracn-Nnleqfracx_N+1+x_N+2+cdots+x_nnleq(L+epsilon)fracn-Nn,
$$
and those bounds tend to $L-epsilon$ and $L+epsilon$ as $ntoinfty$. (Taking the average of a bunch of terms that are close to $L$ should yield a result close to $L$.)
Can you use these ingredients to complete a proof of the result? The intuition is all there.
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
The intuition for a correct proof of this fact is as follows:
For any $epsilon>0$, there exists $N$ so that $lvert x_n-Lrvert<epsilon$ for all $ngeq N$. Equivalently, we can break the sequence into two parts:
Some initial segment $x_1,x_2,ldots, x_N-1$ of terms that can be anything (but there is only a fixed number of them), and
A tail $x_N+1,x_N+2,ldots$ of terms that are all close to (read: within $epsilon$ of) $L$.
If you pick some giant $n$, you get
$$
fracx_1+x_2+cdots+x_nn=fracx_1+cdots+x_Nn+fracx_N+1+x_N+2+cdots+x_nn.
$$
The first term has a fixed numerator, so it tends to $0$ as $ntoinfty$. (The average makes those few initial terms meaningless in the long run.) The second term can easily be seen to satisfy
$$
(L-epsilon)fracn-Nnleqfracx_N+1+x_N+2+cdots+x_nnleq(L+epsilon)fracn-Nn,
$$
and those bounds tend to $L-epsilon$ and $L+epsilon$ as $ntoinfty$. (Taking the average of a bunch of terms that are close to $L$ should yield a result close to $L$.)
Can you use these ingredients to complete a proof of the result? The intuition is all there.
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
answered Jul 22 at 0:05
Nick Peterson
25.5k23758
25.5k23758
add a comment |Â
add a comment |Â
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see also this related post
– G Cab
Jul 22 at 0:31