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 $n>N_1$.

  • $\begingroup$ For the first question, the fact that $\sqrt{n+1}\overset{+\infty}{\sim}\sqrt n$ doesn't imply the result since we have for example $n+1\overset{+\infty}{\sim}n$. But you noticed that $x_n =\frac 1{y_n}$. Did you compute $\lim_{n\to +\infty}y_n$? $\endgroup$ – Davide Giraudo Jun 23 '11 at 14:38
  • $\begingroup$ Since $x_ny_n=1$, what is $x_n$....For the second question, what can you say about each bracket? Second one is less than $\epsilon$, while in the first one one $n$ changes... $\endgroup$ – N. S. Jun 23 '11 at 14:38
  • $\begingroup$ For the third question, maybe you should denote the degree of $p$ by $d$ instead of $n$. You can only check the definition for $\varepsilon <1$. Show that for $n\geq N$ we have $\displaystyle |p(x_n)|\leq \sum_{k=1}^d|a_k|\varepsilon$. $\endgroup$ – Davide Giraudo Jun 23 '11 at 14:42
  • $\begingroup$ 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? $\endgroup$ – miracle173 Jun 23 '11 at 14:44

You have made pretty good progress on all three problems.

For problem 1: note that in fact

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


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

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 $\epsilon$.

  • $\begingroup$ So we want to show that $|p(x_n)| \leq \epsilon$ for all $n > N_1$. We know that $|x_n| \leq \epsilon$ for all $n >N$. Can we choose $N_1 = N$? $\endgroup$ – Damien Jun 23 '11 at 15:06
  • $\begingroup$ @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. $\endgroup$ – Pete L. Clark Jun 23 '11 at 15:46

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 $n\to\infty$.

  • $\begingroup$ $y_n$ should be $x_n$, of course. (+1 anyway.) $\endgroup$ – Shai Covo Jun 23 '11 at 14:49
  • 1
    $\begingroup$ Thanks, Shai. I fixed the x-rated problem. y-not? $\endgroup$ – ncmathsadist Jun 23 '11 at 15:08

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