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Where $a_n = (1 - a_{n-1})(.1)$ when n is even, $= (1 - a_{n-1})(.2)$ when n is odd and $a_1 = .3$. Thank you.

RHS
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2 Answers2

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When $n$ is even you have

$$\begin{align*} a_n&=(1-a_{n-1})(0.1)\\ &=\big(1-(1-a_{n-2})(0.2)\big)(0.1)\\ &=(1-0.2+0.2a_{n-2})(0.1)\\ &=(0.8+0.2a_{n-2})(0.1)\\ &=0.08+0.02a_{n-2}\;, \end{align*}$$

and when $n$ is odd you have

$$\begin{align*} a_n&=(1-a_{n-1})(0.2)\\ &=\big(1-(1-a_{n-2})(0.1)\big)(0.2)\\ &=(1-0.1+0.1a_{n-2})(0.2)\\ &=(0.9+0.1a_{n-2})(0.2)\\ &=0.18+0.02a_{n-2}\;. \end{align*}$$

It’s not hard to check that all terms are positive, so these results show that $a_n\ge0.08$ for each $n\in\Bbb Z^+$, and therefore $\sum_{n\ge 1}a_n$ diverges.

Brian M. Scott
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we must have $lim ^{a_n}_{n \to \infty}=0$

suppose $lim ^{a_n}_{n \to \infty}=l $

for even n we have $a_n=(1-a_{n-1})(.1)$ so

$lim ^{a_n}_{n \to \infty}=lim^{(1-a_{n-1})(.1)}_{n \to \infty}$ so

$10l=1-l \to l=\frac{1}{11}$

then sumation is not convergence

Somaye
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  • May I know more about this method? How do you know the limit of a_n exists? – RHS May 08 '13 at 12:22
  • frist suppose that summation is convergence then we will have contradiction becuase if summation is convergence then we must have lim a_n exist and equal 0 but you can see at the bove it will not happend (lim a_n=0 will not happend) – Somaye May 08 '13 at 12:27
  • Why the only case of convergence is $a_n \to 0$? AND Why the limit of $a_n$ is contradicted when you can find it's limit $l$? – RHS May 10 '13 at 02:50