Simplifying an infinite sum
Question:
The sum of $$1-frac16+frac16timesfrac14-frac16timesfrac14timesfrac{5}{18}+cdots$$
is:
A) $frac23$
B)$frac{2}{sqrt3}$ C)$sqrtfrac23$ D)$frac{sqrt3}{2}$
After looking at the options I thought factoring into two's and three's would be a reasonable approach. The $frac{5}{18}$ I considered factoring first as $frac{2+3}{2times3^2}$ and then as $frac{3^2-2^2}{2times3^2}$, but unfortunately I couldn't find a discernible pattern in either case. Any pointers in the right direction would be appreciated.
sequences-and-series summation
asked Nov 29 at 9:27
s0ulr3aper07
625
|
show 1 more comment
Question:
The sum of $$1-frac16+frac16timesfrac14-frac16timesfrac14timesfrac{5}{18}+cdots$$
is:
A) $frac23$
B)$frac{2}{sqrt3}$ C)$sqrtfrac23$ D)$frac{sqrt3}{2}$
After looking at the options I thought factoring into two's and three's would be a reasonable approach. The $frac{5}{18}$ I considered factoring first as $frac{2+3}{2times3^2}$ and then as $frac{3^2-2^2}{2times3^2}$, but unfortunately I couldn't find a discernible pattern in either case. Any pointers in the right direction would be appreciated.
sequences-and-series summation
asked Nov 29 at 9:27
s0ulr3aper07
625
1
Hint: $frac14 = frac{3}{12}$. The $n$-th term is $(-1)^nfrac{(2n-1)!!}{6^n n!} = left(-frac{1}{12}right)^n binom{2n}{n}$. $binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum.
– achille hui
Nov 29 at 9:43
2
The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1over6}={5over6}=0.833333ldots$. This rules out the first three options (e.g., $sqrt{2/3}=0.816496ldots$), leaving only $sqrt3/2=0.866025ldots$ as a possibly correct answer.
– Barry Cipra
Nov 29 at 9:44
1
Use find $$left(1+dfrac13right)^{-1/2}$$
– lab bhattacharjee
Nov 29 at 9:52
1
@s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence
– lab bhattacharjee
Nov 29 at 10:35
1
Also keep in mind efunda.com/math/exp_log/series_exp.cfm
– lab bhattacharjee
Nov 29 at 10:38
|
show 1 more comment
Question:
The sum of $$1-frac16+frac16timesfrac14-frac16timesfrac14timesfrac{5}{18}+cdots$$
is:
A) $frac23$
B)$frac{2}{sqrt3}$ C)$sqrtfrac23$ D)$frac{sqrt3}{2}$
After looking at the options I thought factoring into two's and three's would be a reasonable approach. The $frac{5}{18}$ I considered factoring first as $frac{2+3}{2times3^2}$ and then as $frac{3^2-2^2}{2times3^2}$, but unfortunately I couldn't find a discernible pattern in either case. Any pointers in the right direction would be appreciated.
sequences-and-series summation
asked Nov 29 at 9:27
s0ulr3aper07
625
Question:
The sum of $$1-frac16+frac16timesfrac14-frac16timesfrac14timesfrac{5}{18}+cdots$$
is:
A) $frac23$
B)$frac{2}{sqrt3}$ C)$sqrtfrac23$ D)$frac{sqrt3}{2}$
After looking at the options I thought factoring into two's and three's would be a reasonable approach. The $frac{5}{18}$ I considered factoring first as $frac{2+3}{2times3^2}$ and then as $frac{3^2-2^2}{2times3^2}$, but unfortunately I couldn't find a discernible pattern in either case. Any pointers in the right direction would be appreciated.
sequences-and-series summation
sequences-and-series summation
asked Nov 29 at 9:27
s0ulr3aper07
625
asked Nov 29 at 9:27
s0ulr3aper07
625
asked Nov 29 at 9:27
s0ulr3aper07
625
asked Nov 29 at 9:27
s0ulr3aper07
625
asked Nov 29 at 9:27
s0ulr3aper07
625
625
1
Hint: $frac14 = frac{3}{12}$. The $n$-th term is $(-1)^nfrac{(2n-1)!!}{6^n n!} = left(-frac{1}{12}right)^n binom{2n}{n}$. $binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum.
– achille hui
Nov 29 at 9:43
2
The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1over6}={5over6}=0.833333ldots$. This rules out the first three options (e.g., $sqrt{2/3}=0.816496ldots$), leaving only $sqrt3/2=0.866025ldots$ as a possibly correct answer.
– Barry Cipra
Nov 29 at 9:44
1
Use find $$left(1+dfrac13right)^{-1/2}$$
– lab bhattacharjee
Nov 29 at 9:52
1
@s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence
– lab bhattacharjee
Nov 29 at 10:35
1
Also keep in mind efunda.com/math/exp_log/series_exp.cfm
– lab bhattacharjee
Nov 29 at 10:38
|
show 1 more comment
1
Hint: $frac14 = frac{3}{12}$. The $n$-th term is $(-1)^nfrac{(2n-1)!!}{6^n n!} = left(-frac{1}{12}right)^n binom{2n}{n}$. $binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum.
– achille hui
Nov 29 at 9:43
2
The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1over6}={5over6}=0.833333ldots$. This rules out the first three options (e.g., $sqrt{2/3}=0.816496ldots$), leaving only $sqrt3/2=0.866025ldots$ as a possibly correct answer.
– Barry Cipra
Nov 29 at 9:44
1
Use find $$left(1+dfrac13right)^{-1/2}$$
– lab bhattacharjee
Nov 29 at 9:52
1
@s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence
– lab bhattacharjee
Nov 29 at 10:35
1
Also keep in mind efunda.com/math/exp_log/series_exp.cfm
– lab bhattacharjee
Nov 29 at 10:38
1
1
Hint: $frac14 = frac{3}{12}$. The $n$-th term is $(-1)^nfrac{(2n-1)!!}{6^n n!} = left(-frac{1}{12}right)^n binom{2n}{n}$. $binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum.
– achille hui
Nov 29 at 9:43
Hint: $frac14 = frac{3}{12}$. The $n$-th term is $(-1)^nfrac{(2n-1)!!}{6^n n!} = left(-frac{1}{12}right)^n binom{2n}{n}$. $binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum.
– achille hui
Nov 29 at 9:43
2
2
The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1over6}={5over6}=0.833333ldots$. This rules out the first three options (e.g., $sqrt{2/3}=0.816496ldots$), leaving only $sqrt3/2=0.866025ldots$ as a possibly correct answer.
– Barry Cipra
Nov 29 at 9:44
The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1over6}={5over6}=0.833333ldots$. This rules out the first three options (e.g., $sqrt{2/3}=0.816496ldots$), leaving only $sqrt3/2=0.866025ldots$ as a possibly correct answer.
– Barry Cipra
Nov 29 at 9:44
1
1
Use find $$left(1+dfrac13right)^{-1/2}$$
– lab bhattacharjee
Nov 29 at 9:52
Use find $$left(1+dfrac13right)^{-1/2}$$
– lab bhattacharjee
Nov 29 at 9:52
1
1
@s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence
– lab bhattacharjee
Nov 29 at 10:35
@s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence
– lab bhattacharjee
Nov 29 at 10:35
1
1
Also keep in mind efunda.com/math/exp_log/series_exp.cfm
– lab bhattacharjee
Nov 29 at 10:38
Also keep in mind efunda.com/math/exp_log/series_exp.cfm
– lab bhattacharjee
Nov 29 at 10:38
|
show 1 more comment
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1
Hint: $frac14 = frac{3}{12}$. The $n$-th term is $(-1)^nfrac{(2n-1)!!}{6^n n!} = left(-frac{1}{12}right)^n binom{2n}{n}$. $binom{2n}{n}$ is known as the central binomial coefficient, look at its wiki entry and you will know how to compute the sum.
– achille hui
Nov 29 at 9:43
2
The pattern is completely unclear (to me, at least), but it does suggest an alternating sum of decreasing terms, from which we can infer that the answer is less than $1$ but greater than $1-{1over6}={5over6}=0.833333ldots$. This rules out the first three options (e.g., $sqrt{2/3}=0.816496ldots$), leaving only $sqrt3/2=0.866025ldots$ as a possibly correct answer.
– Barry Cipra
Nov 29 at 9:44
1
Use find $$left(1+dfrac13right)^{-1/2}$$
– lab bhattacharjee
Nov 29 at 9:52
1
@s0ulr3aper07, In math.stackexchange.com/questions/746388/…, find en.wikipedia.org/wiki/Binomial_series#Convergence
– lab bhattacharjee
Nov 29 at 10:35
1
Also keep in mind efunda.com/math/exp_log/series_exp.cfm
– lab bhattacharjee
Nov 29 at 10:38