# How to prove that $\lim_{n \to \infty} \frac{3^n-1}{2 \cdot 3^n} = \frac{1}{2}$?

I used this limit as an argument in a proof I wrote (Proof by induction that $\sum\limits_{k=1}^n \frac{1}{3^k}$ converges to $\frac{1}{2}$).

I was told I should "prove" the limit but given no indication as to how to go about it. I didn't even know it was possible to formally prove a limit, but if it is I'd love to know how to do it.

\begin{equation*} \frac{3^{n}-1}{2\cdot 3^{n}}=\frac{3^{n}}{2\cdot 3^{n}}-\frac{1}{2\cdot 3^{n}% }=\frac{1}{2}-\frac{1/2}{3^{n}} \end{equation*} Since \begin{equation*} \lim_{n\rightarrow \infty }\frac{1/2}{3^{n}}=0 \end{equation*} then \begin{equation*} \lim_{n\rightarrow \infty }\frac{3^{n}-1}{2\cdot 3^{n}}=\frac{1}{2}-0=\frac{1% }{2}. \end{equation*}

• Ah I see, so you break down the limit in chunks that you can solve. Very cool, thanks. – jeremy radcliff Jun 9 '15 at 2:00

$$\lim_{n \to \infty} \frac{3^n-1}{2 \cdot 3^n} =\lim_{n \to \infty} \frac{\frac{3^n}{3^n} - \frac{1}{3^n}}{\frac{2\cdot 3^n}{3^n}} = \frac{1}{2}$$

This fraction can be written as a sum of two limits: $$\lim \frac{3^n-1}{2\cdot3^n}=\lim \frac{3^n}{2\cdot3^n}-\lim \frac{1}{2\cdot3^n}=\lim \frac{1}{2}-\lim \frac{1}{2\cdot3^n}=\frac{1}{2}-0=\frac{1}{2}$$

Split the limit up

$\lim_{x \rightarrow \infty} \dfrac{3^{n}}{2\cdot 3^{n}} - \dfrac{1}{2\cdot 3^n} = \dfrac{1}{2} - 0$

It is pretty clear after splitting up the fraction.

$$\lim_{n\rightarrow\infty}\frac{3^n-1}{2\cdot 3^n}=\frac{1}{2}\lim_{n\rightarrow\infty}\frac{3^n}{3^n}-\frac{1}{3^n}=\frac{1}{2}(1-0)=\frac{1}{2}$$

Hi it's another way to calculate $S=\Sigma_{i=1}^{\infty}\frac{1}{3^i}$

Just multiply 3 at both side, then you will obtain

$3S=\Sigma_{i=1}^{\infty}3*\frac{1}{3^i} = 1+\Sigma_{i=1}^{\infty}\frac{1}{3^i}$

Do substraction, easy to get $S=\frac{1}{2}$