1. ## 2 Sequence Problems

Find the value of the limit of the sequence defined by:
a(sub 1)=1
a (sub n+1)=13-(1/a(sub n))

I have no idea how to do that.

Find the value of the limit of the sequence defined by:
a(sub 1) = Square root of 10
a(sub n+1) = Square root of (10+a(sub n))

This one either. These are the only two out of 30 that I couldn't do. He said they'd be on the test so I need to learn... bummer.

2. Originally Posted by thegame189
Find the value of the limit of the sequence defined by:
a(sub 1)=1
a (sub n+1)=13-(1/a(sub n))

I have no idea how to do that.

Find the value of the limit of the sequence defined by:
a(sub 1) = Square root of 10
a(sub n+1) = Square root of (10+a(sub n))

This one either. These are the only two out of 30 that I couldn't do. He said they'd be on the test so I need to learn... bummer.
Try this...

For the last one...

suppose $\displaystyle \lim_{n \to \infty}a_n=L$ then $\displaystyle \lim_{n \to \infty}a_{n+1}=L$

$\displaystyle \lim_{n \to \infty}[a_{n+1}=\sqrt{10+a_n}] \iff L=\sqrt{10+L}$

Now we can solve this equation for L

$\displaystyle L=\sqrt{10+L} \iff L^2=10+L \iff L^2-L-10=0$

$\displaystyle L=\frac{1}{2}(1+\sqrt{41})$

3. Ah, I had a sign wrong that I didn't catch. Thanks a bunch!

Any idea how to do the first one?

4. Originally Posted by TheEmptySet
Try this...

For the last one...

suppose $\displaystyle \lim_{n \to \infty}a_n=L$ then $\displaystyle \lim_{n \to \infty}a_{n+1}=L$

$\displaystyle \lim_{n \to \infty}[a_{n+1}=\sqrt{10+a_n}] \iff L=\sqrt{10+L}$

Now we can solve this equation for L

$\displaystyle L=\sqrt{10+L} \iff L^2=10+L \iff L^2-L-10=0$

$\displaystyle L=\frac{1}{2}(1+\sqrt{41})$
heeeem...I'm also having problem with these sequences! We don't know yet that the limit exists! The first steps are to prove that the sequence is bounded(a bit messy for me) and monotone! Then if it is, comes your step!

5. The first sequence

I'll post the proof that $\displaystyle \lim{a_n}$ exists

First we'll show that:$\displaystyle a_{n+1}>a_n$ if $\displaystyle n\geq{1}$

By induction:

$\displaystyle a_2=12>a_1=1$ is true

Assume $\displaystyle a_{n}>a_{n-1}$ holds then we'll prove that: $\displaystyle a_{n+1}>a_n$

$\displaystyle a_n>a_{n-1}$ then $\displaystyle -a_{n}<-a_{n-1}$
so: $\displaystyle -\frac{1}{a_{n-1}}<-\frac{1}{a_{n}}$ and then: $\displaystyle 13-\frac{1}{a_{n-1}}=a_n<13-\frac{1}{a_{n}}=a_{n+1}$ which completes the first part

$\displaystyle a_{n+1}>a_n>...>a_1=1>0$ so it is a positive sequence and therefore it is bounded by 13 since $\displaystyle a_{n+1}=13-\frac{1}{a_n}<13$

This shows that the sequence is incresing and $\displaystyle a_n<13$ for all $\displaystyle n\geq{1}$, thus it converges to say $\displaystyle L$

LIMIT:
Since $\displaystyle a_{n+1}=13-\frac{1}{a_n}$ taking the limit on both sides: $\displaystyle \lim{a_{n+1}}=13-\frac{1}{\lim{a_n}}$ then $\displaystyle L=13-\frac{1}{L}$ now find the roots of this equation and remember that $\displaystyle 13\geq{L}>12=a_2$ to see which one is the limit.