Question

What is a series like below called? I don't know how to approach it...
(Apologies for too many parentheses and any typos, I'm not sure how to write this clearly in this website)
1st term:
1/(1+(1/(2+1)))
2nd term:
1/(1+(1/(2+(1/(1+(1/(2+1))))))
3rd term:
1/(1+(1/(2+(1/(1+(1/(2+(1/(1+1/(2+1))))))))))

In each successive term the last "1" turns into another 1/(1+1/(2+1))

Answer 1

continued fraction?

Answer 2

Is there a way to see if it approaches a particular limit as you add successive terms?

Answer 3

\[\frac{1}{1+\frac{1}{3}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}\]
is that the first one? it is hard for me to read

Answer 4

Yes, that's the first one.
The next one ends up 11/15, then 41/56, then 153/209, if my napkin-math is right.

Answer 5

gimmick is to call the limit \(x\) and then see if you can write an expression in \(x\)

Answer 6

Mm, I played around with that some, but didn't see how to do it.

Answer 7

\[ x= \frac{1}{1+\frac{1}{2+x}} \]

Answer 8

Ah, that looks simple. Thank you both!

Answer 9

Hrm... is there a way to give you both credit for helping? I only see a "best response".

Answer 10

I hope it makes sense... x= continued fraction forever.
replace the part of it with the continued fraction..

Answer 11

no. But sat needs the medals.

Answer 12

try
\[x=\frac{1}{1+\frac{1}{2+x}}\] but maybe this is wrong, not clear

Answer 13

It does make sense - so I can just look at how the fraction behaves with really large values of x. And I'll be happy to give the medal to Sat, then.

Answer 14

oops what @phi said
yeah, i need the medals, going to buy my car with them!

Answer 15

@sat yes, that works....once you solve you can check against 153/209

Answer 16

I'm playing around with an EE question, and wandered into this. Thank you much!

Answer 17

you solve for x
x= -1±sqrt(3)

Answer 18

That looks right. Awesome.