# Thread: Sequence Problem with Parenthesis

1. ## Sequence Problem with Parenthesis

$a_{n} = (\dfrac{n+1}{2n})(1 - \dfrac{1}{n})$

Convergent or divergent?

$\lim n\to \infty[(\dfrac{n+1}{2n})(1 - \dfrac{1}{n})]$

$[(\dfrac{\infty+1}{2\infty})(1 - \dfrac{1}{\infty})]$

$[$(indeterminate) $(1)]$ = indeterminate

Lost on this one. I know the book says the answer is (convergent to 1/2). Do we expand this expression before taking a limit?

2. ## Re: Sequence Problem with Parenthesis

No, the limit of a product is equal to the product of the limits.

Notice \displaystyle \begin{align*} \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n} \to \frac{1}{2} \end{align*} as \displaystyle \begin{align*} n \to \infty \end{align*}.

Also \displaystyle \begin{align*} 1 - \frac{1}{n} \to 1 \end{align*} as \displaystyle \begin{align*} n \to \infty \end{align*}

Thus \displaystyle \begin{align*} \left( \frac{n + 1}{2n} \right) \left( 1 - \frac{1}{n} \right) \to \frac{1}{2} \cdot 1 = \frac{1}{2} \end{align*} as \displaystyle \begin{align*}n \to \infty \end{align*}.

3. ## Re: Sequence Problem with Parenthesis

Originally Posted by Prove It
No, the limit of a product is equal to the product of the limits.

Notice \displaystyle \begin{align*} \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n} \to \frac{1}{2} \end{align*} as \displaystyle \begin{align*} n \to \infty \end{align*}.

Also \displaystyle \begin{align*} 1 - \frac{1}{n} \to 1 \end{align*} as \displaystyle \begin{align*} n \to \infty \end{align*}

Thus \displaystyle \begin{align*} \left( \frac{n + 1}{2n} \right) \left( 1 - \frac{1}{n} \right) \to \frac{1}{2} \cdot 1 = \frac{1}{2} \end{align*} as \displaystyle \begin{align*}n \to \infty \end{align*}.
Ok sounds great. I get it.