How Can I calculate this expression?

Question

I have this repeating expression $5+\dfrac {6} {5+\dfrac {6} {5+..}}$

I saw a solution on a book. which is:

$5+\dfrac {6} {5+\dfrac {6} {5+..}}=x$

$5+\dfrac {6} {x}=x$

$x^2-5x-6=0$

$x=6 $ or $x=-1$ and says $x\ge0$ so $x=6$. is this true? I dont understand why $x$ cant be negative. i.e we know:

$\sum _{n=1}^{\infty }n=-\dfrac {1} {12}$

my second quest. if it is true can we proof the general form $a+\dfrac {b} {a+\dfrac {b} {a+..}}=x\to a+\dfrac {b} {x}=x$

Answer 1

The key is in the dots. It is tough to get a handle on an infinite thing, so usually, it would be a process:

1. Start with $x_1=5$
2. Calculate $x_2=5+6/x_1$
3. Calculate $x_3=5+6/x_2$

Carry on; what is the limit of $x_n$?
Suppose $x_n=6+y_n$. Then $$y_{n+1}=x_{n+1}-6\\ =6/x_n-1\\ =6/(6+y_n)-1\\ =-y_n/(6+y_n)$$ Since $-1\leq y_1\leq1$, we have $|y_{n+1}|\leq|y_n|/5$, so $y_n\to0$ and $x_n\to6$. For general $a,b$, you should do something similar to find which root the process approaches.

Answer 2

I really dislike the equality $$\sum _{n=1}^{\infty }n=-\frac {1} {12}.$$ It is much better to write it as $$\zeta(-1)=-\frac1{12},$$ where $\zeta$ is Riemann's zeta function. And $$\sum _{n=1}^{\infty }n=+\infty$$ is absolutely true.

And for the original question, the method is valid only when the expression has a value (i.e. a certain sequence has a limit). You can prove it by monotone convergence, but I'm not sure you are required to do so, since it seems that the solution is not from a book on calculus, and the convergence is acquiescent.

Answer 3

take 5+6/[5+6/5...]=x then we can write , [5+6/x]=x 5x+6=x^2 x^2-5x-6=0 (1) x=-1 and x=6 are the roots of (1) since the sequence of partial sum of above series is of positvie terms limit cannot be -1[because sequence of +ve terms must converge to a limit >0] hence limit must be equals to '6' hence above sum is equals to '6'