Proving for all n that $\sum_{i=0}^n \frac1{2^{i}} < 2$ 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.
 A: 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
$$ 2 + \frac{1}{2^n} > 2 \iff2 > 1 -\frac{1}{2^n} = \sum_{i=1}^n \frac{1}{2^i}$$
A: 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.
A: $\displaystyle \sum_{k=0}^n \dfrac{1}{2^k} = \dfrac{1 - 2^{-n-1}}{1 - \frac{1}{2}} = 2 - 2^{-n} < 2$
A: 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 $S=\sum_{n=0}^{\infty}\frac{1}{2^n}=\lim_{k\rightarrow\infty} S _k$.
