Can we Combine two convergent infinite series as follows












1












$begingroup$


Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis







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$endgroup$








  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12
















1












$begingroup$


Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12














1












1








1





$begingroup$


Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis







share|cite|improve this question









$endgroup$




Given two infinite series $S_0 = sum_{n=1}^{infty}frac{1}{F(x, n)^{p}}$ and $S_1 = sum_{n=1}^{infty}frac{1}{F'(x, n)^{p}}$.



Let both of these series be 'convergent' to the same real value $K$ for some values of $x$.



Assuming for $x = x_0$ this is the case. Thus,



$S_0 = sum_{n=1}^{infty}frac{1}{F(x_0, n)^{p}} = K$



$S_1 = sum_{n=1}^{infty}frac{1}{F'(x_0, n)^{p}} = K$



Now, can we claim that for new series used by combining the corresponding terms of $S_0$ and $S_1$ are also convergent, i.e.:



$S_2 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) - (frac{1}{F'(x_0, n)^{p}}) = 0$



$S_3 = sum_{n=1}^{infty}(frac{1}{F(x_0, n)^{p}}) + (frac{1}{F'(x_0, n)^{p}}) = 2K$



I assume this is true, but I am not certain.






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 14 '18 at 8:05









TheoryQuest1TheoryQuest1

1498




1498








  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12














  • 1




    $begingroup$
    If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
    $endgroup$
    – RRL
    Dec 14 '18 at 8:12








1




1




$begingroup$
If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
$endgroup$
– RRL
Dec 14 '18 at 8:12




$begingroup$
If $S_n to S $ and $T_n to T$ then $S_n - T_n to S -T$. Letting these denote your partial sum sequences we see its OK.
$endgroup$
– RRL
Dec 14 '18 at 8:12










1 Answer
1






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2












$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18











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1 Answer
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1 Answer
1






active

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active

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active

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2












$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18
















2












$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18














2












2








2





$begingroup$

Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$






share|cite|improve this answer









$endgroup$



Yes, since your $x_0$ is fixed anyway we can just use the standard rule of adding two convergent series:
$$
sum_{n=1}^infty (a_n + b_n) = sum_{n=1}^infty a_n + sum_{n=1}^infty b_n.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 14 '18 at 8:11









BigbearZzzBigbearZzz

8,88821652




8,88821652












  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18


















  • $begingroup$
    Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:16






  • 2




    $begingroup$
    For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
    $endgroup$
    – BigbearZzz
    Dec 14 '18 at 8:17










  • $begingroup$
    Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
    $endgroup$
    – TheoryQuest1
    Dec 14 '18 at 8:18
















$begingroup$
Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:16




$begingroup$
Thank you. I assume by extension its also true for multiplying and division too? And also were $x$ is a complex number too ?
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:16




2




2




$begingroup$
For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:17




$begingroup$
For division you'd need that the divisor doesn't converge to $0$. It also works for complex number. The proofs of these facts are pretty standard and shouldn't be too hard to find.
$endgroup$
– BigbearZzz
Dec 14 '18 at 8:17












$begingroup$
Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:18




$begingroup$
Thanks again. I am just beginning in analysis as a self-interest. my background is CS.
$endgroup$
– TheoryQuest1
Dec 14 '18 at 8:18


















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