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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?


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Yes, sorry about that. – Alti Nov 11 '12 at 7:06
Do you know the result that a bounded monotone increasing sequence of real numbers converges? – Brian M. Scott Nov 11 '12 at 7:10
Yes. So I showed that the sequence of partial sums is bounded. And since $a_n\ge 0$ for all n, then the sequence is monotonically increasing? – Alti Nov 11 '12 at 7:15
Yes. However, what you’ve written here doesn’t actually show that the sequence of partial sums is bounded. Hang on a bit, and I’ll write something up. – Brian M. Scott Nov 11 '12 at 7:18
You’ve shown that $S_1,S_2$, and $S_3$ are less than $1$, but you’ve not done so in a way that obviously generalizes to prove that all $S_n\le 1$. – Brian M. Scott Nov 11 '12 at 7:22
up vote 1 down vote accepted

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


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Much better approach, thank you for all your help – Alti Nov 11 '12 at 7:35
@Alti: You’re welcome. – Brian M. Scott Nov 11 '12 at 7:37

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.

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