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Proving for all n $\in \mathbb N$, $$\sum_{i=0}^n \frac1{2^{i}} < 2$$

Hint. First prove that the left hand side can be expressed in closed form, i.e. without using the summation operator.

This is one of my review questions and I do not know how to start even with the hint. Any help would be appreciated.

Martin Sleziak
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CloudN9ne
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    Hint: sum the term on the left using the geometric series. – Millardo Peacecraft Apr 24 '14 at 03:47
  • Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Apr 24 '14 at 05:43
  • This question is of similar nature: http://math.stackexchange.com/questions/340087/how-would-you-prove-sum-i-1n-3-4i-1-by-induction – Martin Sleziak Apr 24 '14 at 06:18

4 Answers4

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Notice for $|x| < 1 $

$$ 1 + x + x^2 + x^3 + .... + x^r = \frac{1 - x^{r+1}}{1-x} $$

Hence

$$ \sum_{i=1}^n \frac{1}{2^i} = 1 + \left( \frac{1}{2}\right)^1 + \left( \frac{1}{2}\right)^2 + .... + \left( \frac{1}{2}\right)^n = \frac{1 - \left( \frac{1}{2}\right)^{n+1}}{1 - \frac{1}{2}} = 2 - \frac{1}{2^n} $$

And since $\frac{1}{2^n} > 0$ for all $n>0$, then

$$ 2 + \frac{1}{2^n} > 2 \iff2 > 1 -\frac{1}{2^n} = \sum_{i=1}^n \frac{1}{2^i}$$

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We know from elementary school that the sum of the first $n$ terms of a geometric series is $S_n=\frac{a_1(q^n-1)}{q-1}$ where $a_1$ is the first element and $q$ is the common ratio.

Lemma: the sum of a converging (meaning $q<1$ ) infinite geometric series is $S_{\infty}=\frac{a_1}{1-q}$

Proof $S_{\infty} = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a_1(q^n-1)}{q-1}$

Since $q<1$ we get that $\lim_{n \to \infty} q^n=0$ and so:

$S_{\infty} =\lim_{n \to \infty} \frac{a_1(q^n-1)}{q-1} = \frac{a_1(0-1)}{q-1}=\frac{-a_1}{q-1}=\frac{a_1}{1-q}$

Now back to our question: $$\sum_{i=0}^n\frac{1}{2^i} \leq \lim_{k \to \infty} \sum_{i=0}^k\frac{1}{2^i} =\frac{1}{1-\frac{1}{2}}=\frac{1}{\frac{1}{2}}=2$$

Note: the first inequality holds since every element in the series provided is positive.

Oria Gruber
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$\displaystyle \sum_{k=0}^n \dfrac{1}{2^k} = \dfrac{1 - 2^{-n-1}}{1 - \frac{1}{2}} = 2 - 2^{-n} < 2$

DeepSea
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Since $$\sum_{n=0}^{\infty}\frac{1}{2^n}=2$$ (prove this!), and all terms are strictly positive, it must be that any finite sum is strictly less than the infinite series.

In other words, the partial sums $S_k$ form a sequence which strictly increases to its limit, so $$S_0<S_1<S_2<S_3<\cdots <S=2$$ where $S_k=\sum_{n=0}^k \frac{1}{2^n}$ and $S=\sum_{n=0}^{\infty}\frac{1}{2^n}=\lim_{k\rightarrow\infty} S _k$.

MPW
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