nth root of (n + sqrt(n))

**dannyboycurtis** · October 1st 2009, 11:17 AM

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

**qspeechc** · October 1st 2009, 11:32 AM

Originally Posted by dannyboycurtis

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

If you know $\sqrt[n]{n}\rightarrow 1$ doesn't that help you with $\sqrt[n]{2n}\rightarrow 1$ ? If you can show $\sqrt[n]{2}\rightarrow 1$ you'll be done right? You will use the product of convergent sequences theorem.

**Plato** · October 1st 2009, 12:26 PM

Originally Posted by dannyboycurtis

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

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

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