TIFR CSE 2014 | Part A | Question: 16

Question

Let $x_{0}=1$ and

$x_{n+1}= \frac{3+2x_{n}}{3+x_{n}}, n\geq 0$.

$x_{\infty}=\displaystyle \lim_{n\rightarrow \infty}x_{n}$ is

• A. $\left(\sqrt{5}-1\right) / 2$
• B. $\left(\sqrt{5}+1\right) / 2$
• C. $\left(\sqrt{13}-1\right) / 2$
• D. $\left(-\sqrt{13}-1\right) / 2$
• E. None of the above

Answer 1

Answer : C

$x_{n+1}=1+\dfrac{x(n)}{3+x(n)}$

As $n$ tends to infinity, $x_{n+1} = x_n = x$

$x=1+\dfrac{x}{3+x}$

$\implies x^2+3x=3+2x$

$\implies x^2 + x - 3 = 0$

The roots are: $\dfrac{-1+ \sqrt{13}} {2}, \dfrac{-1 - \sqrt{13}} {2}$

Since, $x_n$ is positive,

$x = \dfrac{-1+\sqrt {13}}{2}$

Comments

As nn tends to infinity, xn+1=xn=x. How? I don’t understand this step. Can anyone explain this.

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@neel, we’re assuming here that the series “converges”.

That is, as we go to higher and higher terms of the series, the series doesn’t not explode to infinity, but rather approaches a constant value.

Now that is only possible if for very high numbers, the nth and (n+1) term in the series are very close to each other.

 

For example, consider the sum series 1 + ½ + ¼ + …

The value of this approaches 2.

1, 1.5, 1.75, 1.875, 1.9375, 1.96875, 1.984375, 1.9921875, 1.99609375, 1.998046875, 1.9990234375, 1.99951171875, 1.999755859375, 1.9998779296875, 1.99993896484375, 1.999969482421875, 1.9999847412109375

As you look at higher and higher terms of the series, they all are almost equal to 2.

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Thanks for the explanation. Still, how can we assume that a series converges? We must have to apply some test of convergence to check if a series is convergent or not, right? If so, which test to use?

Answer 2

clearly x(i) >= 1; now dividing numerator and denominator of x(n+1) by x(n)

x(n+1)= (3(/x(n))+2 )/(3/x(n)+1)

but as n increases, 1/(x(n) approaches 0;

therefore limit n-> infinity, x(n+1)=(0+2)/(0+1)=2.

Comments

You're saying that if $n$ is very large, $x_n$ is very large, and so $\frac1{x_n}$ tends to $0$,

But then you said that if $n$ is very large (tends to infinity), then $x_{n+1}$ is very close to $2$. This also means that $x_n$ is very close to $2$. But that contradicts your earlier statement about $x_n$ being very large.

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sorry 2 is wrong

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Is there any alternative way to solve this question?!