I've got a simple question. Can I fill in $\lim_{n \to \infty} n = \lim_{n \to \infty} n+1$ in any equation?
Can I always use that $\lim_{n \to \infty} n = \lim_{n \to \infty} n+1$?
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limits
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4You probably mean "can I always trade $n$ for $n+1$ in a limit to infinity ?". (What you wrote is undefined.) The answer is yes. Make sure to replace all occurrences of $n$. – 2017-01-03
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0Yes, to the question as literally asked. Note however that the replacement itself must be justified separately. For example, it is *not* valid to substitute $\lim (n+1) - n = \lim (n+1) - \lim n = \lim n - \lim n = 0$. – 2017-01-03
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0@dxiv I actually wanted to do what you just noted that I can't. How can I prove that? I'm trying to show that $\lim_{n \to \infty} \sqrt{n}- \sqrt{n+1}= 0$ – 2017-01-03
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0@fatalError $\lim a_n+b_n = \lim a_n + \lim b_n$ holds if both limits on the RHS exist. In the example from my previous comment, they don't. In your case they don't, either, since both terms have $+\infty$ limit. Hint: $\sqrt{n}-\sqrt{n+1}=-1/(\sqrt{n}+\sqrt{n+1})$. – 2017-01-03
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Generalizing your question, $$\lim_{n \to \infty} n = \lim_{n \to \infty} n+k$$ translates to $$\lim_{n \to \infty} n = \lim_{n \to \infty} n(1+k/n)$$ as long as $k$ is small in magnitude with respect to $n$ and can be approximated to zero, you can use either forms.