How can we prove that $\lim_{n\rightarrow\infty}a_{n}=0$ if $a_n>0$ and $\lim_{n\to\infty} a_n / a_{n+1} = l >1$?
Showing that $a_n \to 0$ if a_n/a_{n+1} \to l > 1.
3 Answers
There is some $N$ sufficiently large so that, for all $n > N$, we have $a_{n+1}/a_n \leq c < 1$.
For for $n > 0$ we have $0 < a_{n+N} \leq a_N c^n$. By the squeeze theorem, $a_{n+N} \rightarrow 0$.
-
0I have get the point.thanks – 2012-05-24
Hint:
If $\lim\limits_{n\rightarrow\infty}{a_n\over a_{n+1}}=l>1$, then $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n }={1\over l}<1$. Choose $0
Then note:
$\ \ \ \ a_{N+1}<{c}\, a_N$,
$\ \ \ \ a_{N+2}<{c}\,a_{N+1}<{c^2}a_N$,
$\ \ \ \ a_{N+3}<{c}\,a_{N+2}<{c^3}a_N$,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$
-
0thanks. your step-by-step induction helps me a lot. – 2012-05-24
Another proof not-by-the-definition: the series $\,\displaystyle{\sum_{n=1}^\infty a_n\,}$ is a positive one and it converges by D'Alembert's test (or the quotient test), since $\,\displaystyle{\frac{a_{n+1}}{a_n}\to \frac{1}{l}<1}\,$ , thus it must be $\,a_n\to 0$
-
0No, or at least not the proof I know, which proves the sequence of partial sums of the series is bounded (using a geometric convergente series) and thus the series converges. – 2012-05-24