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Tui a sequence $(a_n)$ defined for all natural numbers given by $$a_1 =1, 2a_{n+1}a_n = 4a_n + 3a_{n+1}, \forall n \geq 1$$

Find the closed formula for the sequence and hence find the limit. Here, what I have done: $$2a_{n+1}a_n = 4a_n + 3a_{n+1} \implies a_{n+1} = \frac{4a_n} {2a_n - 3} \implies a_{n+1} = \frac{\frac{4a_n} {a_n} } {\frac{2a_n}{a_n} - \frac{3} {a_n} } \implies a_{n+1} = \frac{4} {2 - \frac{3} {a_n} } \implies \frac{1 } {a_{n+1}} = \frac{2 - \frac{3} {a_n} } {4} \implies \frac{1 } {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}$$ Then go to where????

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    $\begingroup$ @Salahamam_ Fatima Can you roll the edits back to where I had it, or otherwise break up the lines a little more? Also you broke my improvements on OP's grammar :P $\endgroup$ – Chris Jun 12 '17 at 2:01
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    $\begingroup$ 1. prove that $a_n\ne0$ for all $n$; 2. let $b_n=\frac{(-3)^n}{4^n a_n}$ and find the recursion fomula. $\endgroup$ – Display Name Jun 12 '17 at 2:02
  • $\begingroup$ I can't understand, what you say. $\endgroup$ – Bapon Das Jun 12 '17 at 2:02
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hint

If $a_n=0$ then $a_{n+1}=0$.

assume that for $n>0 \;\;a_n\ne 0$.

dividing by $2a_na_{n+1} $, we get

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

Put $b_n=2/a_n $. then

$$b_{n+1}=1-\frac {3}{4}b_n $$ Now look for $p$ such that

$$b_{n+1}-p=-3/4(b_n-p) $$ thus $$p=4/7$$

$$b_n=4/7+(-3/4)^{n-1}(b_1-4/7) $$ $$=4/7+10/7 (-3/4)^{n-1} $$

$$a_n=\frac {7}{2+5 (-3/4)^{n-1}} $$ $$\lim_{n\to+\infty}=7/2$$

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  • $\begingroup$ $$b_{n+1} - 1/4 = -3/4(b_n-1) $$ $\endgroup$ – MeetR Jun 12 '17 at 3:07
  • $\begingroup$ @MeetR Yes thanks. is it ok now. $\endgroup$ – hamam_Abdallah Jun 12 '17 at 3:22
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So you got $\frac{1} {a_{n+1}} =\frac{1 } {2 } - \frac{3} {4a_n}, a_1=1.$

Denote: $b_n=\frac{1}{a_n}$, then the equation becomes $b_{n+1}=-\frac{3}{4}b_n+\frac12$ with $b_1=\frac{1}{a_1}=1$.

Now apply the general formula: $$\bbox[lightgreen] { If \ \ y_{n+1}=ay_n+b,y_1=c, \ then: y_n=\left(c+\frac{b}{a-1}\right)a^{n-1}-\frac{b}{a-1}. \qquad } $$

Thus:

$$b_n=\left(1+\frac{\frac12}{-\frac34-1}\right)\left(-\frac34\right)^{n-1}-\frac{\frac12}{-\frac34-1}=\frac{5}{7}\left(-\frac34\right)^{n-1}+\frac27$$ and $$a_n=\frac{1}{b_n}=\frac{7}{2+5\left(-\frac34\right)^n}$$

And the limit: $$\lim_\limits{n\to+\infty} a_n=\frac72.$$

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  • $\begingroup$ Thank you all, for giving me answers $\endgroup$ – Bapon Das Jun 13 '17 at 8:16
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Enforcing $u_n=1/a_n$ gives $$2a_{n+1}a_n = 4a_n + 3a_{n+1}\Longleftrightarrow 4u_{n+1}+ 3u_n =2\Longleftrightarrow u_{n+1}=- \frac34u_n +\frac12$$ Now set $x_n = u_n-\frac27$ then you can easily check that we have $x_{n}= -\frac34x_{n-1}$ which is a geometric sequence

$$x_{n}= -\frac34x_{n-1}\implies x_n =\left(-\frac{3}{4}\right)^{n-1}x_1 \\\implies \color{red}{\frac{1}{a_n}= u_n =\frac27+\left(-\frac{3}{4}\right)^{n-1}(u_1 -\frac27)\to \frac27}$$

Alternatively, a more general formula is obtained using this How to find the closed form of $u_{n+1}=a_nu_n+b_n~~~\text{where $a_1$, $a_n$ and $b_n$ are given.}$ you get that $$ u_n = \left(\sum_{p=1}^{n-1}\frac{ \frac12}{\prod_{k=1}^{p}- \frac34}+u_1\right)\prod_{k=1}^{n-1}- \frac34 = \left(-\frac{3}{4}\right)^{n-1}\left(1+\frac12\sum_{p=1}^{n-1}\left(-\frac{4}{3}\right)^{p-1} \right)\\=\color{red}{\frac{1}{a_n}= u_n =\frac27+\frac57\left(-\frac{3}{4}\right)^{n-1} \to \frac27}$$

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If you have proved that the sequence has a limit $l$, then $l$ can be calculated by the following:

$$\lim_{n\to \infty}2a_{n+1}a_n = \lim_{n\to \infty}(4a_n + 3a_{n+1}) $$

and hence

$$2l^2=4l+3l$$

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  • $\begingroup$ I found the limiting value, but how to showw it is convergent. $\endgroup$ – Bapon Das Jun 12 '17 at 2:17
  • $\begingroup$ @BaponDas by its explicit expression $\endgroup$ – hamam_Abdallah Jun 12 '17 at 3:29

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