Mixing
Series - $sum_{i=1}^infty (frac{5}{12})^i$ - geometric series?
$begingroup$
I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?
The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$
However in my assignment, the series starts from $i=1$.
The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$
Can you explain please why is that the solution?
calculus sequences-and-series taylor-expansion
edited Jan 16 at 14:04
Larry
2,41331129
asked Jan 16 at 13:57
AlanAlan
1,3841021
$endgroup$
add a comment |
$begingroup$
I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?
The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$
However in my assignment, the series starts from $i=1$.
The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$
Can you explain please why is that the solution?
calculus sequences-and-series taylor-expansion
edited Jan 16 at 14:04
Larry
2,41331129
asked Jan 16 at 13:57
AlanAlan
1,3841021
$endgroup$
4
$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57
add a comment |
$begingroup$
I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?
The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$
However in my assignment, the series starts from $i=1$.
The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$
Can you explain please why is that the solution?
calculus sequences-and-series taylor-expansion
edited Jan 16 at 14:04
Larry
2,41331129
asked Jan 16 at 13:57
AlanAlan
1,3841021
$endgroup$
I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?
The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$
However in my assignment, the series starts from $i=1$.
The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$
Can you explain please why is that the solution?
calculus sequences-and-series taylor-expansion
calculus sequences-and-series taylor-expansion
edited Jan 16 at 14:04
Larry
2,41331129
asked Jan 16 at 13:57
AlanAlan
1,3841021
edited Jan 16 at 14:04
Larry
2,41331129
asked Jan 16 at 13:57
AlanAlan
1,3841021
edited Jan 16 at 14:04
Larry
2,41331129
edited Jan 16 at 14:04
Larry
2,41331129
edited Jan 16 at 14:04
Larry
2,41331129
2,41331129
asked Jan 16 at 13:57
AlanAlan
1,3841021
asked Jan 16 at 13:57
AlanAlan
1,3841021
asked Jan 16 at 13:57
AlanAlan
1,3841021
1,3841021
4
$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57
add a comment |
4
$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57
4
4
$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57
$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
$endgroup$
3
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
1
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
add a comment |
$begingroup$
There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$
answered Jan 16 at 14:20
BernardBernard
121k740116
$endgroup$
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
add a comment |
$begingroup$
Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.
answered Jan 16 at 14:40
drhabdrhab
102k545136
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
$endgroup$
3
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
1
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
add a comment |
$begingroup$
HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
$endgroup$
3
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
1
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
add a comment |
$begingroup$
HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
$endgroup$
HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
answered Jan 16 at 13:59
Ahmad BazziAhmad Bazzi
8,3282824
8,3282824
3
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
1
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
add a comment |
3
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
1
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
3
3
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
$endgroup$
– Alan
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
exactly correct @Alan
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:01
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
Thank you for your answer and for your time.
$endgroup$
– Alan
Jan 16 at 14:02
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
$begingroup$
no problem @Alan .. you can mark the answer as correct if you found it useful :)
$endgroup$
– Ahmad Bazzi
Jan 16 at 14:03
1
1
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
$begingroup$
Yes, in a few minutes when I can :-)
$endgroup$
– Alan
Jan 16 at 14:03
add a comment |
$begingroup$
There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$
answered Jan 16 at 14:20
BernardBernard
121k740116
$endgroup$
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
add a comment |
$begingroup$
There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$
answered Jan 16 at 14:20
BernardBernard
121k740116
$endgroup$
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
add a comment |
$begingroup$
There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$
answered Jan 16 at 14:20
BernardBernard
121k740116
$endgroup$
There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$
answered Jan 16 at 14:20
BernardBernard
121k740116
answered Jan 16 at 14:20
BernardBernard
121k740116
answered Jan 16 at 14:20
BernardBernard
121k740116
answered Jan 16 at 14:20
BernardBernard
121k740116
121k740116
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
add a comment |
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
$begingroup$
Thank you, a great point of view.
$endgroup$
– Alan
Jan 16 at 14:23
add a comment |
$begingroup$
Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.
answered Jan 16 at 14:40
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.
answered Jan 16 at 14:40
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.
answered Jan 16 at 14:40
drhabdrhab
102k545136
$endgroup$
Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.
answered Jan 16 at 14:40
drhabdrhab
102k545136
answered Jan 16 at 14:40
drhabdrhab
102k545136
answered Jan 16 at 14:40
drhabdrhab
102k545136
answered Jan 16 at 14:40
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
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4
$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57