Evaluate 1 + 3/4 + (3*5)/(4*8)+…
$begingroup$
Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...
I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.
I am not able to to relate this sum with Taylor expansion of some function
real-analysis taylor-expansion
asked Jan 9 at 4:45
Swapnil Swapnil
415
$endgroup$
add a comment |
$begingroup$
Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...
I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.
I am not able to to relate this sum with Taylor expansion of some function
real-analysis taylor-expansion
asked Jan 9 at 4:45
Swapnil Swapnil
415
$endgroup$
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17
add a comment |
$begingroup$
Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...
I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.
I am not able to to relate this sum with Taylor expansion of some function
real-analysis taylor-expansion
asked Jan 9 at 4:45
Swapnil Swapnil
415
$endgroup$
Evaluate
1 + 3/4 + (3.5)/(4.8) + (3.5.7)/(4.8.12) +...
I have simplified it to
Summation ((2k+1)!)/(k!)(k!)) (1/8)^k where k varies from 0 to infinity.
I am not able to to relate this sum with Taylor expansion of some function
real-analysis taylor-expansion
real-analysis taylor-expansion
asked Jan 9 at 4:45
Swapnil Swapnil
415
asked Jan 9 at 4:45
Swapnil Swapnil
415
asked Jan 9 at 4:45
Swapnil Swapnil
415
asked Jan 9 at 4:45
Swapnil Swapnil
415
asked Jan 9 at 4:45
Swapnil Swapnil
415
415
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17
add a comment |
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17
$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Sounds like you are looking for
$$
sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
$$
so consider differentiating
$$
f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
$$
and using $(1/8)^k = (1/sqrt8)^{2k}$
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
$endgroup$
add a comment |
$begingroup$
the series is the binomial expansion of
$$
bigg(1 - frac12 bigg)^{-frac32}
$$
which gives the sum $2sqrt{2}$
edited Jan 9 at 5:42
Ross Millikan
297k23198371
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
$endgroup$
1
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sounds like you are looking for
$$
sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
$$
so consider differentiating
$$
f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
$$
and using $(1/8)^k = (1/sqrt8)^{2k}$
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
$endgroup$
add a comment |
$begingroup$
Sounds like you are looking for
$$
sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
$$
so consider differentiating
$$
f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
$$
and using $(1/8)^k = (1/sqrt8)^{2k}$
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
$endgroup$
add a comment |
$begingroup$
Sounds like you are looking for
$$
sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
$$
so consider differentiating
$$
f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
$$
and using $(1/8)^k = (1/sqrt8)^{2k}$
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
$endgroup$
Sounds like you are looking for
$$
sum_{k = 0}^infty (2k+1) binom{2k}{k}left( frac18 right)^k,
$$
so consider differentiating
$$
f(x) = sum_{k = 0}^infty binom{2k}{k} x^{2k+1},
$$
and using $(1/8)^k = (1/sqrt8)^{2k}$
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
answered Jan 9 at 4:57
gt6989bgt6989b
34.3k22455
34.3k22455
add a comment |
add a comment |
$begingroup$
the series is the binomial expansion of
$$
bigg(1 - frac12 bigg)^{-frac32}
$$
which gives the sum $2sqrt{2}$
edited Jan 9 at 5:42
Ross Millikan
297k23198371
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
$endgroup$
1
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
add a comment |
$begingroup$
the series is the binomial expansion of
$$
bigg(1 - frac12 bigg)^{-frac32}
$$
which gives the sum $2sqrt{2}$
edited Jan 9 at 5:42
Ross Millikan
297k23198371
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
$endgroup$
1
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
add a comment |
$begingroup$
the series is the binomial expansion of
$$
bigg(1 - frac12 bigg)^{-frac32}
$$
which gives the sum $2sqrt{2}$
edited Jan 9 at 5:42
Ross Millikan
297k23198371
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
$endgroup$
the series is the binomial expansion of
$$
bigg(1 - frac12 bigg)^{-frac32}
$$
which gives the sum $2sqrt{2}$
edited Jan 9 at 5:42
Ross Millikan
297k23198371
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
edited Jan 9 at 5:42
Ross Millikan
297k23198371
edited Jan 9 at 5:42
Ross Millikan
297k23198371
edited Jan 9 at 5:42
Ross Millikan
297k23198371
297k23198371
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
answered Jan 9 at 5:27
David HoldenDavid Holden
14.9k21224
14.9k21224
1
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
add a comment |
1
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
1
1
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
$begingroup$
MathJax hint: if you use left and right before parentheses and other delimiters they autosize based on what is inside.
$endgroup$
– Ross Millikan
Jan 9 at 5:43
add a comment |
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$begingroup$
math.stackexchange.com/questions/746388/…
$endgroup$
– lab bhattacharjee
Jan 9 at 5:17