Prove that $\sum_{k=0}^{n} \frac{1}{4^{k}}=\frac{4}{3}-\frac{1}{3 \cdot 4^{n}}$ converges to $\frac{4}{3}$ using the $\varepsilon-N$ proof

Solution 1:

Let $a_n := \tfrac43 - \tfrac1{3 \cdot 4^n}$. Now, you want to show that $S_n \to \tfrac43$. Given $\varepsilon > 0$, you desire $|a_n - \tfrac43| \le \varepsilon$. But the left-hand side is precisely $1/(3 \cdot 4^n)$. So just choose the smallest $N$ such that $1/(3 \cdot 4^N) \le \varepsilon$, ie $N := \lceil \log_4(\tfrac1{3\varepsilon}) \rceil$, as you rightly noted.

Solution 2:

I think we could take a not so strange norm for $N$. As you well know, it all comes down to proving that $\frac{1}{3\cdot 4^n}\to 0$.

Let $\epsilon>0$, we can take $N=\left\lfloor\dfrac{1}{\epsilon}\right\rfloor+1>\dfrac{1}{\epsilon}$, then for $n>N$ we have $$ \left|\frac{1}{3\cdot 4^n}\right|<\left|\frac{1}{3\cdot 4^N}\right|<\frac{1}{N}<\epsilon $$ Thus $\frac{4}{3}-\frac{1}{3\cdot 4^n}\to \frac{4}{3}$ when $n\to \infty$.