# Cauchy/convergent sequence

• Oct 17th 2011, 08:22 PM
limddavid
Cauchy/convergent sequence
Question: Let {a_n} be a sequence of rational numbers defined as follows:

a_1 = 1
a_(n+1)=a_n + 1/(3^n)

for all n > 1.

1) Show that {a_n} is a Cauchy sequence and hence convergent
2) Find it's limit.

So.. I am a bit lost for this one. (Headbang) I tried using induction to bound a_n by 1 and 2, but I couldn't bound it above. I figured if I bound it above, and prove that it's a non-decreasing function, then I could prove that it converges, and thus that it is Cauchy...

Any tips for me to get started on this? Thanks!
• Oct 17th 2011, 08:38 PM
chisigma
Re: Cauchy/convergent sequence
Quote:

Originally Posted by limddavid
Question: Let {a_n} be a sequence of rational numbers defined as follows:

a_1 = 1
a_(n+1)=a_n + 1/(3^n)

for all n > 1.

1) Show that {a_n} is a Cauchy sequence and hence convergent
2) Find it's limit.

So.. I am a bit lost for this one. (Headbang) I tried using induction to bound a_n by 1 and 2, but I couldn't bound it above. I figured if I bound it above, and prove that it's a non-decreasing function, then I could prove that it converges, and thus that it is Cauchy...

Any tips for me to get started on this? Thanks!

The difference equation can be written as...

$\Delta_{n}= a_{n+1}-a_{n}= \frac{1}{3^{n}}\ ,\ a_{1}=1$ (1)

... and its solution is...

$a_{n}= a_{1} + \sum_{k=1}^{n-1} \Delta_{k}= a_{1} + \sum_{k=1}^{n-1} \frac{1}{3^{k}}$ (2)

From (1) and (2) You derive that is...

$\lim_{n \rightarrow \infty} a_{n}= 1 + \sum_{k=1}^{\infty} \frac{1}{3^{k}}= 1 + \frac{1}{3\ (1-\frac{1}{3})}= 1+\frac{1}{2}= \frac{3}{2}$ (3)

Kind regards

$\chi$ $\sigma$
• Oct 17th 2011, 08:41 PM
limddavid
Re: Cauchy/convergent sequence
riight! sum of geometric series. I'm not 1000% sure I can use this in my class right now, but it's worth a try. thank you!