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I have to prove for $a_n$\in${0,1}, $\sum_{n=1}^\infty \frac{a_n}{2^n} always converges for all $n\inℕ$.

I took the extreme examples, where the sequence is either all zeroes or all ones. If $a_n$ is a sequence of zeroes, then $S_n$ (sequence of partial sums) will be zero. If $a_n$ is a sequence of ones, then

S_1=\frac12$ $S_2=\frac12+\frac14$ $S_3=\frac12+\frac14+\frac18$$ and on. Therefore, in this case, $S_n\le\frac12+\frac12=1

So 0\le S_n\le1$

Is this a correct approach?

Thanks

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    You can use induction, or you can do what I did in my answer. Let me know if you want me to add the inductive argument.2012-11-11

2 Answers 2

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To show that the sequence of partial sums is bounded in the extreme case of the constant $1$ sequence, just use the formula for the sum of a finite geometric progression, which I’ve actually derived here:

Let $S_n=\sum_{k=1}^n\frac1{2^k}\;;$ then

$\begin{align*} \frac12S_n&=\frac12\sum_{k=1}^n\frac1{2^k}=\sum_{k=1}^n\frac1{2^{k+1}}\\ &=\sum_{k=2}^{n+1}\frac1{2^k}=\left(\sum_{k=1}^{n+1}\frac1{2^k}\right)-\frac12\\ &=\left(\sum_{k=1}^n\frac1{2^k}\right)+\frac1{2^{n+1}}-\frac12\\ &=S_n+\frac1{2^{n+1}}-\frac12\;. \end{align*}$

Now solve for $S_n$:

$\frac12S_n=\frac12-\frac1{2^{n+1}}\;,$ so $S_n=1-\frac1{2^n}<1\;.$

Now for the general case you have

$\sum_{k=1}^n\frac{a_k}{2^k}\le\sum_{k=1}^n\frac1{2^k}<1\;.$

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    @Alti: You’re welcome.2012-11-11
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Given that the sequence of $b_n=\frac{a_n}{2^n}$ is nonnegative, it suffices to find a dominating sequence whose series converges. In particular, $b_n\leq \frac1{2^n}$ for all $n$, so since the geometric series $\sum_{n=1}^\infty\frac1{2^n}$ converges (to $1$), then the series $\sum_{n=1}^\infty b_n$ also converges.