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$a_1=1 \,\text{and} \,$$a_{n+1}=(1-\frac{1}{2^n})a_n$

My try:This is a decreasing sequence and bounded below by $1$.So $a_n$ converges.

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    $\begingroup$ How is this bounded below by 1 while $a_2=\frac 1 2 $? $\endgroup$ – Shashi Feb 21 '17 at 18:47
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The first elements of the sequence:

$$a_1=1,\,a_2=\frac12,\,a_3=\frac38,\,a_4=\frac{21}{64}\,,\ldots$$

so it looks like a descending sequence. With a little induction

$$a_{n+1}:=\left(1-\frac1{2^n}\right)a_n\le a_n\iff1-\frac1{2^n}\le1$$

and since the last inequality is trivial we're done.

Finally, again with a little induction

$$a_{n+1}=\left(1-\frac1{2^n}\right)a_n\ge0\iff a_n\ge0$$

and thus zero is a lower bound, and thus the sequnece converges.

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You are correct except the lower bound should not be 1. A lower bound of 0 works though, so by the monotone convergence theorem the sequence converges.

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Check that,

$$a_n = \prod_{k=1}^{n-1}\frac{a_{k+1}}{a_k} = \prod_{k=1}^{n-1}(1-\frac{1}{2^k}) $$

Therefore $$\lim_{n\to \infty}\ln a_n = \sum_{n=0}^\infty \ln(1-\frac{1}{2^n})$$ which converges since

$$\lim_{n\to \infty} 2^n \ln(1-\frac{1}{2^n}) = \lim_{h\to 0} \frac{\ln(1-h)}{h} = 1$$ that is $$\ln(1-\frac{1}{2^n}) \approx \frac{1}{2^n} $$

$\lim\limits_{n\to \infty}\ln a_n$ exists then $\lim_{n\to \infty} a_n$ exists

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