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Prove that a_n tends to 0

Alexmahone

Active member
Jan 26, 2012
268
Assume $\frac{a_{n+1}}{a_n}\to L$, where $L<1$ and $a_n>0$. Prove that

(a) $\{a_n\}$ is decreasing for $n\gg 1$;

I've done this part.

(b) $a_n\to 0$ (Give two proofs: an indirect one using (a), and a direct one.)

Indirect proof: Assume for the sake of argument that $a_n$ does not tend to 0. Since the sequence is decreasing and bounded below by 0, it must converge to a positive value (call it $l$).

$\lim\frac{a_{n+1}}{a_n}=\frac{l}{l}=1$, a contradiction. So, $a_n\to 0$.

How do I go about the direct proof (one that doesn't use "proof by contradiction")?
 
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Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
How do I go about the direct proof (one that doesn't use "proof by contradiction")?
Hint

$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.
 

Alexmahone

Active member
Jan 26, 2012
268
Hint

$|a_{n}-0|<\epsilon\Leftrightarrow |a_{n+1}\cdot (a_n/a_{n+1})|<\epsilon \Leftrightarrow |a_{n+1}|<|a_{n+1}/a_n|\epsilon$ and $a_{n+1}/a_n$ is bounded.
Is this a proof by induction that $|a_n|<\epsilon$? If so, how do we prove the base case for $n=1$?
 
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HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.
 

Alexmahone

Active member
Jan 26, 2012
268
I would, rather, prove by induction that $0< a_n< L^n a_1$ and show that that geometric sequence converges to 0.
I used a slightly different approach:

$L<\frac{L+1}{2}$

$\frac{a_{n+1}}{a_n}<\frac{L+1}{2}$ for $n\gg 1$ (Using the "sequence location theorem")

So, $a_{n+1}< \left(\frac{L+1}{2}\right)a_n$ for $n\ge N$

$a_{N+k}<\left(\frac{L+1}{2}\right)^k a_N$ for $k\ge 1$ (Can be proved using induction over $k$.)

Since $\frac{L+1}{2}<1$, $a_{N+k}\to 0$ as $k\to\infty$.

So, $a_n\to 0$.
 
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