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I understand the concept of convergence and am comfortable doing proofs with the definition. I am also adept at basic proofs such as "Prove that $\lim\limits_{n\to\infty} (\frac{1}{\sqrt{n}})=0$." However, this question is stumping me; in particular, I'm having a hard time justifying a particular choice of $N$.

Typically, when doing simpler problems, the choice of $N$ is somewhat readily apparent. I'm usually able to identify a function of $n$ which is always less than $\epsilon$, and then manipulate the sequence $(a_n)$ into that form to evaluate the limit. For example, if I were to prove that $\lim\limits_{n\to\infty} (\frac{n+1}{n})=1$, I would use the fact that, given any $\epsilon>0$, I can choose $N$ such that $\frac{1}{n}<\epsilon$ whenever $n\ge{N}$. Then, I simply use the fact that $|\frac{n+1}{n}-1|=\frac1n<\epsilon$, and I'm basically done.

However, with this particular problem, I'm having difficulty choosing $N$, and I'm honestly not quite sure where to start since the sequence is a rather complicated one (although I may be missing something very obvious). The problem mentions that I should do this using just the definition of convergence if that's any help. Any pointers would be helpful. Please let me know if I can clarify anything since English is not my native language.

Awnon Bhowmik
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salamanderman
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    A good first step would be to "multiply by the conjugate", similar to the way you clear square roots out of denominators. – JonathanZ Jul 04 '23 at 18:15
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    It is NOT duplicate, since the linked post is NOT answering this OP's question on how to choose $N$. Vote to re-open. – MathFail Jul 04 '23 at 18:35
  • Here is another duplicate target: https://math.stackexchange.com/q/783536/42969 (including “epsilon-N proofs”) – and this has been asked and answered frequently: https://math.stackexchange.com/questions/linked/783536 – Martin R Jul 04 '23 at 18:42

2 Answers2

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$$\left|\sqrt{n^2+n}-n-\frac12 \right|=\left|\frac{n^2+n-(n+\frac12)^2}{\sqrt{n^2+n}+n+\frac12}\right|=\frac14\cdot\frac{1}{\sqrt{n^2+n}+n+\frac12}\le \frac1{n}<\epsilon$$

Hence, you can choose $N=\frac1\epsilon$

MathFail
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Let $n\geq n_{\varepsilon}$. Then one has: \begin{align*} \left|\sqrt{n^{2} + n} - n - \frac{1}{2}\right| & = \left|\frac{n}{\sqrt{n^{2} + n} + n} - \frac{1}{2}\right|\\\\ & \leq \frac{\sqrt{n^{2} + n} - n}{\sqrt{n^{2} + n} + n}\\\\ & = \frac{n}{(\sqrt{n^{2} + n} + n)^{2}}\\\\ & \leq \frac{1}{n} \leq \frac{1}{n_{\varepsilon}} < \varepsilon \end{align*}

From then on, I assume you can proceed.

Átila Correia
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