Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across a theroem which says that if $S=(ar^{n-1}-a)/(r-1)$ when $r \neq 1$ then $S=(n+1)a$ if $r=1$. But for $r=1$ the above equation isn't well-defined. How do they come to this result?

share|improve this question
    
Have a look at De l'Hôpital's Theorem. –  Lord_Farin Oct 24 '12 at 14:04
    
It should be $S=(n-1)a,$ or $S=(ar^{n+1}-a)/(r-1)$. –  Kevin Carlson Oct 24 '12 at 15:06
add comment

2 Answers

Be careful: we understand what you mean, but you wrote a rather arbitrary thing. What is true is that $$ \lim_{r \to 1} \frac{ar^{n+1}-a}{r-1}=(n+1)a, \tag{1} $$ but $S$ is undefined at $r=1$. The limit (1) is simply $a$ times the derivative of $r \mapsto r^{n+1}$ at $r=1$.

share|improve this answer
    
using l'Hopital's Rule proofs this easily –  user31280 Oct 24 '12 at 14:26
    
@F'OlaYinka ... but you need a different proof of the formula $\frac{d}{dr}r^{n+1}=(n+1)r^n$ :-) –  Siminore Oct 24 '12 at 14:29
    
it's a polynomial function, very obvious –  user31280 Oct 24 '12 at 14:43
add comment

This is the formula for a geometric progression. When $r\neq 1$, the sequence is $a,ar,ar^2,...$ so that the sum of the first $n$ terms is $\frac{a(r^n-1)}{r-1}$. When $r=1$, the sequence is $a,a,a,...$ so that the sum of the first $n$ terms is $na$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.