I agree with your answer, but your calculations are bogus.
Where does the 6/7 come from ??
r = (1/28) / (1/7) = 1/4
sum(all terms) = a/(1-r) = (1/7) /(3/4) = 4/21
Does the following infinite geometric series diverge or converge? Explain.
1/7+1/28+1/112+1/448
A) It converges; it has a sum.***
B) It diverges; it does not have a sum.
C) It diverges; it has a sum.
D) It converges; it does not have a sum.
Since (1/7)/(6/7)=1/6
and -1>1/6>1
6 answers
I had asked a question similar to this before. You showed a similar way of solving it.
Does the following infinite geometric series diverge or converge? Explain.
1/5 + 1/25 + 1/125 + 1/625
This is what you said
Notice that is simply a geometric series, where
a = 1/5, r = 1/5
since Sum(all terms) = a/(1-r)
= (1/5)/(4/5)
= 1/4 , it clearly converges
Does the following infinite geometric series diverge or converge? Explain.
1/5 + 1/25 + 1/125 + 1/625
This is what you said
Notice that is simply a geometric series, where
a = 1/5, r = 1/5
since Sum(all terms) = a/(1-r)
= (1/5)/(4/5)
= 1/4 , it clearly converges
I also was thinking the same thing about 4/5 when you said this
Every geometric series where |r| < 1
has a sum and thus is converging.
Both examples and problems are like that
the formula for sum(infinite number of terms)
= a/(1-r)
I used that in both cases
In your latest example, a = 1/7 and r = 1/4
sum(all terms) = (1/7) / (1 - 1/4)
= (1/7) / (3/4)
= (1/7)(4/3)
= 4/21 , as I stated above in my first reply
I don't understand what you find confusing.
has a sum and thus is converging.
Both examples and problems are like that
the formula for sum(infinite number of terms)
= a/(1-r)
I used that in both cases
In your latest example, a = 1/7 and r = 1/4
sum(all terms) = (1/7) / (1 - 1/4)
= (1/7) / (3/4)
= (1/7)(4/3)
= 4/21 , as I stated above in my first reply
I don't understand what you find confusing.
I think I am starting to understand it. Am I doing this correctly now.
Does the following infinite geometric series diverge or converge? Explain.
3 + 9 + 27 + 81+...
r=3 a(1st term?)=3
Sum=3/1-3 = 3/-2 = -(3/2) = -1.5
Does the following infinite geometric series diverge or converge? Explain.
3 + 9 + 27 + 81+...
r=3 a(1st term?)=3
Sum=3/1-3 = 3/-2 = -(3/2) = -1.5
NOOO!
For convergence, the common ratio has to be a proper fraction, see above when I said |r| < 1
Is this case the common ratio is 3, so the terms keep getting bigger and bigger
How can the sum zoom in on a specific number?
in the first case, each consecutive term is getting smaller and smaller. So you are adding less and less each time.
1/7+1/28 = 5/28 = appr .1786
1/7+1/28+1/112 = 3/16 = appr .1875
1/7+1/28+1/112+1/448 = 85/448 = appr .1897
adding the next term yields appr .19029
and another yields appr .19042
notice the increase is less and less
my final answer was 4/21 which is appr .19047619
My sum of only the first 6 terms is only off by
.0000465... and we have billions and billions more to go
For convergence, the common ratio has to be a proper fraction, see above when I said |r| < 1
Is this case the common ratio is 3, so the terms keep getting bigger and bigger
How can the sum zoom in on a specific number?
in the first case, each consecutive term is getting smaller and smaller. So you are adding less and less each time.
1/7+1/28 = 5/28 = appr .1786
1/7+1/28+1/112 = 3/16 = appr .1875
1/7+1/28+1/112+1/448 = 85/448 = appr .1897
adding the next term yields appr .19029
and another yields appr .19042
notice the increase is less and less
my final answer was 4/21 which is appr .19047619
My sum of only the first 6 terms is only off by
.0000465... and we have billions and billions more to go
1 answer