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

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

Now we can solve this equation for L

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

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

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

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

Now we can solve this equation for L

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

$L=\frac{1}{2}(1+\sqrt{41})$
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 $\lim{a_n}$ exists

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

By induction:

$a_2=12>a_1=1$ is true

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

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

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

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

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