Question

Use the definition of limit to prove that the sequence { 1/√n+1 } converges to 0

Answer 1

The sequence $\left(\frac{1}{\sqrt{n}+1}\right)_n$ converges to 0 if

$$\forall \varepsilon>0,\;\exists N\in\mathbb{N},\; \left(n\geq N\implies \left|\tfrac{1}{\sqrt{n}+1}-0\right|=\tfrac{1}{\sqrt{n}+1}\leq\varepsilon\right).$$

Let $\varepsilon>0$. If we take $N=\left\lceil\dfrac{1}{\varepsilon}\right\rceil^2$ we obtain $\sqrt{N}=\left\lceil\dfrac{1}{\varepsilon}\right\rceil\geq\dfrac{1}{\varepsilon}$. Then $\dfrac{1}{\sqrt{N}}\leq\varepsilon$. Therefore

$$\frac{1}{\sqrt{N}+1}\leq\frac{1}{\sqrt{N}}\leq\varepsilon.$$

So, for all $n\geq N,$

$$\frac{1}{\sqrt{n}+1}\leq\frac{1}{\sqrt{N}+1}\leq\epsilon$$