Answer to Question #158166 in Calculus for sai

Question #158166
Use the definition of limit to prove that the sequence { 1/√n+1 } converges to 0
1
Expert's answer
2021-01-26T15:09:23-0500

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


ε>0,  NN,  (nN    1n+10=1n+1ε).\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 ε>0\varepsilon>0. If we take N=1ε2N=\left\lceil\dfrac{1}{\varepsilon}\right\rceil^2 we obtain N=1ε1ε\sqrt{N}=\left\lceil\dfrac{1}{\varepsilon}\right\rceil\geq\dfrac{1}{\varepsilon}. Then 1Nε\dfrac{1}{\sqrt{N}}\leq\varepsilon. Therefore

1N+11Nε.\frac{1}{\sqrt{N}+1}\leq\frac{1}{\sqrt{N}}\leq\varepsilon.

So, for all nN,n\geq N,

1n+11N+1ϵ\frac{1}{\sqrt{n}+1}\leq\frac{1}{\sqrt{N}+1}\leq\epsilon


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