Let \(A= 1+\dfrac1 4+ \dfrac{1}{16}+...=\displaystyle \sum_{n=0}^{\infty}\dfrac{1}{4^n}\).

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

\(S_n=\displaystyle \sum_{i=0}^n \dfrac{1}{4^i}=\dfrac{1-\dfrac{1}{4^{n+1}}}{1-\dfrac1 4}=\dfrac4 3 \times \Bigg (1-\dfrac{1}{4^{n+1}} \Bigg )\)

Let \(n \rightarrow \infty\), we have \(S_n \rightarrow A\), and \(S_n \rightarrow \dfrac4 3\).

As the limit of \(S_n\) is unique, we have \(A=\dfrac4 3\).

Let A=1+14+116+...=∑n=0∞14n

.

We calculate the sum of n

first numbers of the series A

.

Sn=∑i=0n14i=1−14n+11−14=43×(1−14n+1)

Let n→∞

, we have Sn→A, and Sn→43

.

As the limit of Sn

is unique, we have A=43.