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I'm fairly confident I got the right idea, but I'm not quite sure how to state the answer...

\begin{align} \lim_{n\to\infty}\frac{1+a+a^2+\dots+a^n}{1+b+b^2+\dots+b^n}&=\frac{\lim_{n\to\infty}1+a+a^2+\dots+a^n}{\lim_{n\to\infty}1+b+b^2+\dots+b^n}\\ &=\frac{\lim_{n\to\infty}\frac{1-a^{n+1}}{1-a}}{\lim_{n\to\infty}\frac{1-b^{n+1}}{1-b}}\\ &=\frac{\lim_{n\to\infty}(1-a^{n+1})(1-b)}{\lim_{n\to\infty}(1-a)(1-b^{n+1})}\\ &=\begin{cases}\frac{1-b}{1-a}&&\text{if } |a|,|b|<1\\\infty&&\text{if }|a|>1,|b|<1\\0&&\text{if }|a|<1,|b|>1\\ DNE&&\text{otherwise}\end{cases} \end{align} I'm not confident on the DNE part, and I don't know if there are other cases I'm missing. Also, fwiw, this is for a real analysis class, where the assignment says "You need to show your calculation, but no need to use '$\epsilon-N$' language to prove."

Mirrana
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  • Well, $|a|=1$ implies $a=1$ or $a=-1$. So your last case splits in 4 cases. – rom Oct 15 '13 at 21:25
  • What if it evaluates to $\infty/\infty$? I was always told that this is not definitive, and might require L'Hopital's rule... It may converge and may not. – Mirrana Oct 15 '13 at 21:29
  • If it evaluates to ${\infty \over \infty}$ use l'Hopital on ${1 - a^{n+1} \over 1- b^{n+1}}$. Also you have to consider the case where $|a| = 1$ or $|b|=1$. – Zarrax Oct 15 '13 at 21:46
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    If $a=b=1$ then obviously the limit is 1. – rom Oct 15 '13 at 21:54
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    l'Hospital cannot be used for multivariable formula in the same way. – Math.StackExchange Oct 15 '13 at 22:17

1 Answers1

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If $a\ge b> 1$,

$\frac{\lim_{n\to\infty}(1-a^{n+1})(1-b)}{\lim_{n\to\infty}(1-a)(1-b^{n+1})}=\lim_{n\to\infty}\frac{1-b}{1-a}\frac{b^{-(n+1)}-(a/b)^{(n+1)}}{b^{-(n+1)}-1}=\infty$

Since $|\frac{\lim_{n\to\infty}1+a+a^2+\dots+a^n}{\lim_{n\to\infty}1-b+b^2+\dots+(-1)^nb^n}|\ge|\frac{\lim_{n\to\infty}1+a+a^2+\dots+a^n}{\lim_{n\to\infty}1+b+b^2+\dots+b^n}|$,

if $a\ge -b>1$,

$\frac{\lim_{n\to\infty}1+a+a^2+\dots+a^n}{\lim_{n\to\infty}1+b+b^2+\dots+b^n}=\infty$

If $b\ge a> 1$,

$\frac{\lim_{n\to\infty}(1-a^{n+1})(1-b)}{\lim_{n\to\infty}(1-a)(1-b^{n+1})}=0$

Since $|\frac{\lim_{n\to\infty}1-a+a^2+\dots+(-1)^na^n}{\lim_{n\to\infty}1+b+b^2+\dots+b^n}|\le|\frac{\lim_{n\to\infty}1+a+a^2+\dots+a^n}{\lim_{n\to\infty}1+b+b^2+\dots+b^n}|$,

if $b\ge -a>1$,

$\frac{\lim_{n\to\infty}1+a+a^2+\dots+a^n}{\lim_{n\to\infty}1+b+b^2+\dots+b^n}=0$

Not sure about other cases.

Math.StackExchange
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