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!

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...

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

... and its solution is...

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

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

$\displaystyle \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

$\displaystyle \chi$ $\displaystyle \sigma$

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!