Verify using the definition of convergence that the following sequences converge to the proposed limit:
- \(\lim\limits_{n \rightarrow \infty} \frac{n^2 + 1}{2n^2 + 3} = \frac{1}{2}\).
- \(\lim\limits_{n \rightarrow \infty} \frac{3^n}{3^n + 2^n} = 1\).
- \(\lim\limits_{n \rightarrow \infty} \frac{\sin n}{n}\). [TODO]
Definitions of sequences and convergence: here, Definitions of subsequences: here.
For the “show the limit” template: here.
Solution
(a) To prove that \(\lim \frac{n^2 + 1}{2n^2 + 3} = \frac{1}{2}\), then we need to show that for \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that when \(n \geq N\), we must have
$$
\begin{align*}
\left\lvert \frac{n^2 + 1}{2n^2 + 3} - \frac{1}{2} \right\rvert &< \epsilon
\end{align*}
$$
We can simplify the left hand side to see that
$$
\begin{align*}
\left\lvert \frac{n^2 + 1}{2n^2 + 3} - \frac{1}{2} \right\rvert &=
\left\lvert \frac{2n^2 + 2}{2(2n^2 + 3)} - \frac{2n^2 + 3}{2(2n^2 + 3)} \right\rvert \\
&=
\left\lvert \frac{2n^2 + 2 - 2n^2 - 3}{2(2n^2 + 3)} \right\rvert \\
&=
\left\lvert \frac{-1}{2(2n^2 + 3)} \right\rvert \\
&=
\frac{1}{4n^2 + 6}
\end{align*}
$$
Solving for \(n\) in terms of \(\epsilon\)
$$
\begin{align*}
\frac{1}{4n^2 + 6} &< \epsilon \\
\frac{1}{\epsilon} &< 4n^2 + 6 \\
\frac{1}{\epsilon} - 6 &< 4n^2 \\
\frac{1}{4}\left(\frac{1}{\epsilon} - 6 \right) &< n^2 \\
\sqrt{\frac{1}{4}\left(\frac{1}{\epsilon} - 6 \right)} &< n
\end{align*}
$$
We want to choose \(n \geq N\) such that the inequality above holds. Observe that when \(\epsilon \geq \frac{1}{6}\), then this inequality holds for any \(n\) so we can set \(N = 1\). Otherwise, we can set \(N\) as follows
$$
\begin{align*}
N = \left\lceil\sqrt{\frac{1}{4}\left(\frac{1}{\epsilon} - 6\right)}\right\rceil
\end{align*}
$$
Thus, for all \(\epsilon > 0\), we found an \(N \in \mathbb{N}\) such that for all \(n \geq N\)
$$
\begin{align*}
\left\lvert \frac{n^2 + 1}{2n^2 + 3} - \frac{1}{2} \right\rvert &< \epsilon
\end{align*}
$$
This implies that
$$
\begin{align*}
\lim \frac{n^2 + 1}{2n^2 + 3} = \frac{1}{2}
\end{align*}
$$
as we wanted to show. \(\blacksquare\)
(b) To prove that \(\lim \frac{3^n}{3^n + 2^n} = 1\), then we need to show that for \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that when \(n \geq N\), we must have
$$
\begin{align*}
\left\lvert \frac{3^n}{3^n + 2^n} - 1 \right\rvert &< \epsilon
\end{align*}
$$
We can simplify the left hand side to see that
$$
\begin{align*}
\frac{3^n}{3^n + 2^n} - 1 &=
\left\lvert \frac{3^n}{3^n + 2^n} - \frac{3^n + 2^n}{3^n + 2^n} \right\rvert \\
&=
\left\lvert \frac{3^n - 3^n - 2^n}{3^n + 2^n} \right\rvert \\
&=
\left\lvert \frac{- 2^n}{3^n + 2^n} \right\rvert \\
&=
\frac{2^n}{3^n + 2^n}
\end{align*}
$$
So we want to bound the following term to be less than \(\epsilon\)
$$
\begin{align*}
\frac{2^n}{3^n + 2^n} &< \epsilon
\end{align*}
$$
But notice that since \(2^n > 0\) and \(3^n > 0\), then
$$
\begin{align*}
\frac{2^n}{3^n + 2^n} &< \frac{2^n}{3^n} = \left( \frac{2}{3}\right)^n
\end{align*}
$$
This is an easier fraction to work with. Thus we want to solve for \(n\) in
$$
\begin{align*}
\left( \frac{2}{3}\right)^n &< \epsilon \\
\log \left( \frac{2}{3}\right)^n &< \log (\epsilon) \\
n\log \left( \frac{2}{3}\right) &< \log (\epsilon) \\
n &> \frac{\log(\epsilon)}{\log\left(\frac{2}{3}\right)}
\end{align*}
$$
Therefore, we can pick \(N\) to be
$$
\begin{align*}
N &> \left\lceil \frac{\log(\epsilon)}{\log\left(\frac{2}{3}\right)} \right\rceil
\end{align*}
$$
Thus, for all \(\epsilon > 0\), we found an \(N \in \mathbb{N}\) such that for all \(n \geq N\)
$$
\begin{align*}
\left\lvert \frac{3^n}{3^n + 2^n} - 1 \right\rvert &< \epsilon
\end{align*}
$$
This implies that
$$
\begin{align*}
\lim \frac{3^n}{3^n + 2^n} = 1
\end{align*}
$$
as we wanted to show. \(\blacksquare\)
References