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Question about Proof of Merten's theorem (Cauchy-Product formula)
$begingroup$
I have a question about the proof on the german wikepedia page:
The proof is stated as follow:
Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.
Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$
Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$
1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$
2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$
1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$
$(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:
$sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$
Then
$|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$
Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$
Hence
$|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$
I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.
Thank you for your time, I would appreciate your help very much.
sequences-and-series analysis
asked Dec 22 '18 at 12:38
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
I have a question about the proof on the german wikepedia page:
The proof is stated as follow:
Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.
Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$
Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$
1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$
2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$
1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$
$(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:
$sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$
Then
$|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$
Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$
Hence
$|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$
I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.
Thank you for your time, I would appreciate your help very much.
sequences-and-series analysis
asked Dec 22 '18 at 12:38
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
I have a question about the proof on the german wikepedia page:
The proof is stated as follow:
Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.
Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$
Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$
1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$
2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$
1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$
$(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:
$sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$
Then
$|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$
Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$
Hence
$|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$
I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.
Thank you for your time, I would appreciate your help very much.
sequences-and-series analysis
asked Dec 22 '18 at 12:38
RM777RM777
38312
$endgroup$
I have a question about the proof on the german wikepedia page:
The proof is stated as follow:
Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.
Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$
Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$
1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$
2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$
1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$
$(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:
$sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$
Then
$|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$
Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$
Hence
$|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$
I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.
Thank you for your time, I would appreciate your help very much.
sequences-and-series analysis
sequences-and-series analysis
asked Dec 22 '18 at 12:38
RM777RM777
38312
asked Dec 22 '18 at 12:38
RM777RM777
38312
edited Feb 1 at 19:32
RM777
asked Dec 22 '18 at 12:38
RM777RM777
38312
asked Dec 22 '18 at 12:38
RM777RM777
38312
asked Dec 22 '18 at 12:38
RM777RM777
38312
38312
add a comment |
add a comment |
1 Answer
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$begingroup$
The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'
Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.
As for the max. The task is to estimate a sum of the form
$sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates
$$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$
answered Feb 1 at 19:48
quid♦quid
37.2k95193
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'
Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.
As for the max. The task is to estimate a sum of the form
$sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates
$$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$
answered Feb 1 at 19:48
quid♦quid
37.2k95193
$endgroup$
add a comment |
$begingroup$
The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'
Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.
As for the max. The task is to estimate a sum of the form
$sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates
$$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$
answered Feb 1 at 19:48
quid♦quid
37.2k95193
$endgroup$
add a comment |
$begingroup$
The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'
Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.
As for the max. The task is to estimate a sum of the form
$sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates
$$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$
answered Feb 1 at 19:48
quid♦quid
37.2k95193
$endgroup$
The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'
Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.
As for the max. The task is to estimate a sum of the form
$sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates
$$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$
answered Feb 1 at 19:48
quid♦quid
37.2k95193
answered Feb 1 at 19:48
quid♦quid
37.2k95193
answered Feb 1 at 19:48
quid♦quid
37.2k95193
answered Feb 1 at 19:48
quid♦quid
37.2k95193
37.2k95193
add a comment |
add a comment |
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