# nth root of (n + sqrt(n))

• Oct 1st 2009, 11:17 AM
dannyboycurtis
nth root of (n + sqrt(n))
I have to show that $\displaystyle \lim\sqrt[n]{n+\sqrt{n}}$ goes to 1 as $\displaystyle n\rightarrow\infty$.
I have that $\displaystyle \sqrt[n]{n}\leq \sqrt[n]{n+\sqrt{n}}\leq\sqrt[n]{2n}$
I know that $\displaystyle \sqrt[n]{n}\rightarrow 1$ via the binomial theorem.
If I can show that $\displaystyle \sqrt[n]{2n}\rightarrow 1$, I could use the Squeeze theorem.
Any ideas?
• Oct 1st 2009, 11:32 AM
qspeechc
Quote:

Originally Posted by dannyboycurtis
I have to show that $\displaystyle \lim\sqrt[n]{n+\sqrt{n}}$ goes to 1 as $\displaystyle n\rightarrow\infty$.
I have that $\displaystyle \sqrt[n]{n}\leq \sqrt[n]{n+\sqrt{n}}\leq\sqrt[n]{2n}$
I know that $\displaystyle \sqrt[n]{n}\rightarrow 1$ via the binomial theorem.
If I can show that $\displaystyle \sqrt[n]{2n}\rightarrow 1$, I could use the Squeeze theorem.
Any ideas?

If you know $\displaystyle \sqrt[n]{n}\rightarrow 1$ doesn't that help you with $\displaystyle \sqrt[n]{2n}\rightarrow 1$? If you can show $\displaystyle \sqrt[n]{2}\rightarrow 1$ you'll be done right? You will use the product of convergent sequences theorem.
• Oct 1st 2009, 12:26 PM
Plato
Quote:

Originally Posted by dannyboycurtis
I have to show that $\displaystyle \lim\sqrt[n]{n+\sqrt{n}}$ goes to 1 as $\displaystyle n\rightarrow\infty$.
I have that $\displaystyle \sqrt[n]{n}\leq \sqrt[n]{n+\sqrt{n}}\leq\sqrt[n]{2n}$
I know that $\displaystyle \sqrt[n]{n}\rightarrow 1$ via the binomial theorem.
If I can show that $\displaystyle \sqrt[n]{2n}\rightarrow 1$, I could use the Squeeze theorem.

Any ideas? Yes: $\displaystyle \sqrt[n]{{2n}} = \sqrt[n]{2}\sqrt[n]{n}$.