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Integral that is continuous and looks like it converges to a geometric series
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Integral that is continuous and looks like it converges to a geometric series
Testing if a geometric series converges by taking limit to infinitySummation of arithmetic-geometric series of higher orderGeometric series with polynomial exponentHow to Recognize a Geometric SeriesShowing an integral equality with series over the integersDiscontinuity of a series of continuous functionsReasons why a Series ConvergesSum of infinite geometric series with two terms in summationUsing geometric series for computing IntegralsLimit of geometric series sum when $r = 1$
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I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
$endgroup$
add a comment |
$begingroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
$endgroup$
add a comment |
$begingroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
$endgroup$
I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .
calculus integration multivariable-calculus improper-integrals
calculus integration multivariable-calculus improper-integrals
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
asked 5 hours ago
Randin DRandin D
1026
1026
add a comment |
add a comment |
1 Answer
1
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votes
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Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
add a comment |
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
add a comment |
$begingroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
$endgroup$
Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
$$h(a) = int_0^1 h(x) , dx.$$
answered 5 hours ago
angryavianangryavian
43k23482
answered 5 hours ago
angryavianangryavian
43k23482
answered 5 hours ago
angryavianangryavian
43k23482
answered 5 hours ago
angryavianangryavian
43k23482
43k23482
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
add a comment |
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
$begingroup$
aha ..do u have an email we can chat more about this problemo?
$endgroup$
– Randin D
5 hours ago
add a comment |
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