Convergence of the Sequence √n(√(n+1)-√n) to 1/2

[Mr Davis 97]

Homework Statement

Show that $\displaystyle \lim_{n\to \infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n}) = \frac{1}{2}$

Homework Equations

The Attempt at a Solution

We see that $\displaystyle \sqrt{n}(\sqrt{n+1}-\sqrt{n}) - \frac{1}{2} = \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{2} < \frac{\sqrt{n}}{2\sqrt{n}} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0$. But I don't think this can be right... What am I doing wrong?

 

[Eclair_de_XII]

$\frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$

Try dividing the top and bottom of this fraction by something, then take the limit as $n\rightarrow \infty$.

 

[fresh_42]

Homework Statement

Show that $\displaystyle \lim_{n\to \infty} \sqrt{n}(\sqrt{n+1}-\sqrt{n}) = \frac{1}{2}$

Homework Equations

The Attempt at a Solution

We see that $\displaystyle \sqrt{n}(\sqrt{n+1}-\sqrt{n}) - \frac{1}{2} = \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{2} < \frac{\sqrt{n}}{2\sqrt{n}} - \frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0$. But I don't think this can be right... What am I doing wrong?

Nothing, except that the step of subtracting $\dfrac{1}{2}$ is unnecessary. You can directly go to
$$
\sqrt{n}(\sqrt{n+1}-\sqrt{n})=\dfrac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}< \dfrac{\sqrt{n}}{2\sqrt{n}}=\dfrac{1}{2}
$$
Now the limit is definitely smaller than a half. What a bout the lower bound?

 

[Mr Davis 97]

Nothing, except that the step of subtracting $\dfrac{1}{2}$ is unnecessary. You can directly go to
$$
\sqrt{n}(\sqrt{n+1}-\sqrt{n})=\dfrac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}< \dfrac{\sqrt{n}}{2\sqrt{n}}=\dfrac{1}{2}
$$
Now the limit is definitely smaller than a half. What a bout the lower bound?

Well a lower bound would be 0. But I am trying to do this using the definition of convergence. I'm doing the scratchwork to find the $N$ and can associate with every $\epsilon$. Like normally in the end we get like a difference between the sequence and the limit as $1/n < \epsilon$, and so we select $N$ s.t. $1/N < \epsilon$, and by the archimdedian principle this exists. However, in this case I just get that the difference is 0...

 

[fresh_42]

Well a lower bound would be 0. But I am trying to do this using the definition of convergence. I'm doing the scratchwork to find the $N$ and can associate with every $\epsilon$. Like normally in the end we get like a difference between the sequence and the limit as $1/n < \epsilon$, and so we select $N$ s.t. $1/N < \epsilon$, and by the archimdedian principle this exists. However, in this case I just get that the difference is 0...

You need to prove something like $\frac{1}{2}-\frac{c}{n} \leq a_n$ so that the gap is closing down with increasing $n$.
However, the easier method is to follow @Eclair_de_XII 's suggestion and simply cancel out $\sqrt{n}$ and then use the rules given for limit arithmetic. My question only refers to the way you have chosen, in which case you need a lower bound.