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Consider the sequence $ \ \large \ a_n=\frac{1}{4^{\left\lfloor \frac{n}{2} \right\rfloor}+(k+1) \ mod \ 2} \ $ .

Find the order of convergence and rate of convergence .

Answer:

From the definition of the sequence the first few terms are $ 1/2,1, 1/5, 1/4, 1/17, 1/16 , 1/65, 1/64 ,....... \ $

Thus

$ \large \lim_{n \to \infty} \frac{a_{n+1}}{a_n} <1 \ $ as well as $ \ \large \lim_{n \to \infty} \frac{a_{n+1}}{a_n} >1 \ $

We know that if $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} <1 \ $ , then the convergence is linear and if $ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} >1 \ $ , then the sequence will be divergent .

I am getting confused here. which one is correct ?

The rate of convergence is given by

$ \large \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \\ =\lim_{n \to \infty} \frac{4^{\left\lfloor \frac{n}{2} \right\rfloor}+(n+1) \ mod \ 2}{4^{\left\lfloor \frac{n+1}{2} \right\rfloor}+(n+2) \ mod \ 2} \\= ? $

I am unable to determine right here also.

Please help me out.

MAS
  • 10,638

1 Answers1

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$$ \frac{a_{n+1}}{a_n} = \frac{4^{\left[{\frac n2}\right]} + (n + 1) \ mod \ 2}{4^{\left[{\frac {n+1}{2}}\right]} + (n + 2) \ mod \ 2} $$

Let $n$ be even.

$$ \frac{a_{n+1}}{a_n} = \frac{4^\frac n2 + 1}{4^\frac n2} = 1 + 4^{-\frac n2} \longrightarrow 1 $$

Let $n$ be odd.

$$ \frac{a_{n+1}}{a_n} = \frac{4^\frac {n-1}{2}}{4^\frac {n-1}{2} + 1} = 1 - \frac{1}{4^\frac {n-1}{2} + 1} \longrightarrow 1 $$

Here we conclude that

$$ \frac{a_{n+1}}{a_n} \longrightarrow 1 \ \mbox{while} \ n \longrightarrow \infty $$

It means that an order of convergence is sublinear.

Elnur
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