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I came across the following problems on null sequences during the course of my self-study of real analysis.

Let $x_n = \sqrt{n+1}- \sqrt{n}$. Is $(x_n)$ a null sequence?

Consider $y_n = \sqrt{n+1}+ \sqrt{n}$. Then $x_{n}y_{n} = 1$ for all $n$. So either $(x_n)$ or $(y_n)$ is not a null sequence. It seems $(y_n)$ is not a null sequence. I think $(x_n)$ is a null sequence because $\sqrt{n+1} \approx \sqrt{n}$ for large $n$ which implies that $x_n \approx 0$ for large $n$.

If $(x_n)$ is a null sequence and $y_n = (x_1+ x_2+ \dots + x_n)/n$ then $(y_n)$ is a null sequence.

Suppose $|x_n| \leq \epsilon$ for all $n >N$. If $n>N$, then $y_n = y_{N}(N/n)+ (x_{N+1}+ \dots+ x_n)/n$. From here what should I do?

If $p: \mathbb{R} \to \mathbb{R}$ is a polynomial function without constant term and $(x_n)$ is a null sequence, then $p((x_n))$ is null.

We know that $|x_n| \leq \epsilon$ for all $n>N$. We want to show that $|p(x_n)| \leq \epsilon$ for all $n>N_1$. We know that $p(x) = a_{d}x^{d} + \cdots+ a_{1}x$. So $|p(x_n)| \leq a_{d} \epsilon^{d} + \cdots+ a_{1} \epsilon$

for all $n>N_1$.

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    I'm not sure about the meaning of $\approx$ but I think your argumentation i wrong. From $x_ny_n=1$ follows $x_n=\frac{1}{y_n}$. Perhaps you can show that $\frac{1}{y_n}$ is a null sequence?2011-06-23

2 Answers 2

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You have made pretty good progress on all three problems.

For problem 1: note that in fact

$\lim_{n \rightarrow \infty} y_n = \lim_{n \rightarrow \infty} \sqrt{n+1} + \sqrt{n} = \infty + \infty = \infty$,

so

$\lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} \frac{1}{y_n} = \frac{1}{\infty} = 0$.

For problem 2: you have

$y_n = y_N (N/n) + (x_{N+1} + \ldots + x_n)/n$.

It's enough to show that both terms of the right hand side can be made arbitrarily small as $n$ gets arbitrarily large. The first term is a constant divided by $n$: this goes to zero with $n$. The second term is a sum of at most $n$ things each one of which is in absolute value at most $\frac{\epsilon}{n}$, so the sum is in absolute value at most $\epsilon$. So you're basically done.

For problem 3: If you choose $\epsilon \leq 1$ then $\epsilon^n \leq \epsilon$ for all $n \geq 1$, so

$|a_d \epsilon^d + \ldots + a_1 \epsilon| \leq |a_d + \ldots + a_1| \epsilon$, a quantity which goes to zero with $\epsilon$.

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    @Damien: no, we cannot. Think about the easy case in which $p(x) = a x$: we will be "off by a constant". In general, we want to choose $N$ such that for all $n \geq N$, we have |a_1 + \ldots + a_d| |x_n| < \epsilon. I am starting to get the impression that you are good at finding the right algebraic manipulations to do these proofs but are a little shaky on the logic behind them. If so, perhaps you could ask further questions with more of a focus in this direction.2011-06-23
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Try this.
$x_n = {(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1} + \sqrt{n})\over \sqrt{n+1} + \sqrt{n}} = {1\over \sqrt{n+1} + \sqrt{n}}.$ This shows $x_n\to 0$ as $n\to\infty$.

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    Thanks, Shai. I fixed the x-rated problem. y-not?2011-06-23