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I have the sequence $(a_{n})_{n}$ with first term $a_{1}=2$ and recursive relation $$a_{n+1}=\frac{2}{1+a_{n}}$$ So I have to find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$

Any help?

Niki Sedlarević
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  • $$a_n=\frac{6}{\left(-\frac{1}{2}\right)^n+2}-2$$ even indices give $$a_{2k}=1-\frac{3}{2^{2 k+1}+1}$$ and odd indices give $$a_{2k+1}=\frac{3}{4^{k+1}-1}+1$$ – Raffaele Feb 05 '21 at 16:39

3 Answers3

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$$a_{n+2}=\frac{2}{1+a_{n+1}}=\frac{2+2a_n}{3+a_{n}}$$

Therefore the 'odd' terms satisfy

$$b_1=a_1=2, b_{n+1}=\frac{2+2b_n}{3+b_{n}}$$

and the 'even' terms satisfy the same relation with $$b_1=a_2=\frac{2}{3}.$$

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Well, "to find" means an explicit expression, doesn't it? So you just prove $$a_n=\frac{2^{n+1}-2\,(-1)^n}{2^{n+1}+(-1)^n}$$ by induction (that's almost trivial, given the recursive relation), and then, you get the subsequences, using $(-1)^{2k}=1$ and $(-1)^{2k+1}=-1$.

This may seem a bit ad hoc, but it's in the spirit of the question (no context, no source, no motivation,...).

NoNames
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Multiplying by the denominator we arrive to $$a_na_{n+1}=2-a_{n+1}$$

In this kind of problem, we want to shift $a_n$ so as to cancel the constant term (see this other problem for instance https://math.stackexchange.com/a/3412108/399263)

If you are not sure about which shift to apply, you can set $b_n=a_n+\alpha$ and work out possible values of $\alpha$.

$(b_n-\alpha)(b_{n+1}-\alpha)=2-(b_{n+1}-\alpha)\iff b_nb_{n+1}=(\alpha-1)b_{n+1}+\alpha\,b_n+\underbrace{(2+\alpha-\alpha^2)}_\text{make it zero}$

We have two possibilities:

  • $\alpha=2$ and set $b_n=a_n+2$
  • $\alpha=-1$ and set $b_n=a_n-1$

Remark that now if we divide by $b_nb_{n+1}\quad$ we get $\quad1=(\alpha-1)\dfrac 1{b_n}+\alpha\dfrac 1{b_{n+1}}$

So let introduce $c_n=\dfrac 1{b_n}$ we arrive at a linear induction equation:

$$\alpha\, c_{n+1}+(\alpha-1)c_n=1$$

Let solve for $\alpha=2$, as an exercise you can try on your own with the other value $-1$ and see if you get the same result.

We get $2c_{n+1}+c_n=1$ so homogeneous equation solves to $(-\frac 12)^nC$ and particular solution is constant $2D+D=1\iff D=\frac 13$

Therefore $c_n=(-\frac 12)^nC+\frac 13$.

We have to solve for initial condition $a_1=2=b_1-2=\dfrac 1{c_1}-2=\dfrac 1{-\frac C2+\frac 13}-2\iff C=\frac 16$

After reporting in the expression of $a_n$ and simplifying we get the result proposed in NoName's and Raffaele's answers.

$$a_n=\dfrac{1}{\frac 16(-\frac 12)^n+\frac 13}-2=\dfrac{2^{n+1}-2(-1)^n}{2^{n+1}+(-1)^n}$$

zwim
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