Let \(A= 1+\dfrac1 2+ \dfrac1 4+...=\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{2^n}\).

We calculate the sum of \(n\) first numbers of the series \(A \).

\(S_n=\displaystyle \sum_{i=0}^n \dfrac{1}{2^i}=\dfrac{1-\dfrac{1}{2^{n+1}}}{1-\dfrac1 2}=2-\dfrac{1}{2^n}\).

Let \(n \rightarrow \infty\), we have \(S_n \rightarrow A\), and \(S_n \rightarrow 2\).

As the limit of \(S_n\) is unique, we have \(A=2\).

Note that the denotation of \(\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{2^n}\) is just a way we write for short, in fact we cannot find the exact sum of the infinite series, we just can find the limit of the sequence \(S_n\) and we define \(\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{2^n}= \lim_{n\rightarrow \infty}S_n\).

Let A=1+12+14+...=∑n=0∞12n

.

We calculate the sum of n

 first numbers of the series A

.

Sn=∑i=0n12i=1−12n+11−12=2−12n

.

Let n→∞

, we have Sn→A, and Sn→2

.

As the limit of Sn

 is unique, we have A=2

.

Note that the denotation of ∑n=0∞12n

 is just a way we write for short, in fact we cannot find the exact sum of the infinite series, we just can find the limit of the sequence Sn and we define ∑n=0∞12n=limn→∞Sn.