a simple issue
- Thread starter sujoy
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I do not know how you "know" that. It makes little sense. Where did you get such an impression?
1) Infinity is not a number
2) Do you mean in the limit? Absolutely not the case.
Try this as n increases without bound (1 + (a/n))<sup>n</sup>.
1) Infinity is not a number
2) Do you mean in the limit? Absolutely not the case.
Try this as n increases without bound (1 + (a/n))<sup>n</sup>.
okay--0kay
the thing is that when we sum up a G.P., series the formula is given as :
S=a(n^r -1)/(r-1)
where s means sum, n is the no. of terms& r is the common ratio
what will happen when r is infinity?
:idea: dont we reach infinity as 1+2+3+-------to infinity, then why it is not a number
thanks
sujoy
the thing is that when we sum up a G.P., series the formula is given as :
S=a(n^r -1)/(r-1)
where s means sum, n is the no. of terms& r is the common ratio
what will happen when r is infinity?
:idea: dont we reach infinity as 1+2+3+-------to infinity, then why it is not a number
thanks
sujoy
- Joined
- Feb 4, 2004
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- 16,583
"S" is not the same as "s", and should not be used interchangeably.
"1 + 2 + 3 + 4 + ..." is not a geometric progression (which is what I'm assuming "G.P." stands for).
The given summation formula only applies when |r| < 1, and does not apply to the series you provided.
"Infinity" is not a number, so neither the common ratio r nor the number of terms n can "be" infinity.
Eliz.
"1 + 2 + 3 + 4 + ..." is not a geometric progression (which is what I'm assuming "G.P." stands for).
The given summation formula only applies when |r| < 1, and does not apply to the series you provided.
"Infinity" is not a number, so neither the common ratio r nor the number of terms n can "be" infinity.
Eliz.
pka
Elite Member
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Please note that the n<sup>th</sup> partial sum is usually written as S<sub>n</sub>=a(1−r<sup>n</sup>)/(1−r) [not n<sup>r</sup> ].
As Eliz pointed out, this sequence converges if and only if |r|<1. In that case,
Lim (r<sup>n</sup>)=0. THUS
n→∞
Lim (S<sub>n</sub>)=(a)/(1-r).
n→∞
As Eliz pointed out, this sequence converges if and only if |r|<1. In that case,
Lim (r<sup>n</sup>)=0. THUS
n→∞
Lim (S<sub>n</sub>)=(a)/(1-r).
n→∞