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How do I show, using only basic methods, that a product
$$a_n=(1+{1 \over 2}) \cdot (1+{1 \over 4}) \cdot ... \cdot(1+{1 \over 2^n})$$ converges?

I know, that $a_1 = {3\over2}$, and that $a_{n+1} = a_n \cdot (1+ {1 \over 2^{n+1}})$.

Going further:
$$ {{a_{n+1}}\over{a_n}} = 1+ {1 \over 2^{n+1}} > 1$$


Now I can prove by induction, that $a_n$ is monotonically increasing, and is greater than one.

Now, I want to prove, that $a_n$ is convergent. How can I do this? I can't seem to find any sequence greater than $a_n$ with a limit other than + infinity.

Edit
Following @Barry Cipra comment, i tried taking $ln(a_n)$, so:
$$ ln(a_n) = ln((1+{1 \over 2}) \cdot (1+{1 \over 4}) \cdot ... \cdot(1+{1 \over 2^n}))$$ $$ ln(a_n) = ln(1+{1 \over 2}) + ln(1+{1 \over 4}) + ... +ln(1+{1 \over 2^n})$$ $$ ln(a_n) < 1+{1 \over 2}+1+ {1 \over 4}+...+1+{1 \over 2^n} = n +1$$ $$ ln(a_n) < n+1 $$ $$ a_n < e^{n+1} $$

But I still don't know how this gets me any closer to the answer

grand_chat
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Karol Szustakowski
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1 Answers1

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It is true that $\ln (x)\le x$ for every $x$, but a sharper inequality is: $$ \ln (x) \le x-1\qquad{\text{for all $x$}}. $$ Using this inequality you get $$\ln (a_n) \le \sum_{k=1}^n \frac1{2^k}\le 1, $$ which implies that $(a_n)$ is bounded above. From this, along with what you've already established, we deduce that $(a_n)$ converges.

grand_chat
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