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

Suppose $a_n \to a$. Define $s_n = \frac{1}{n}\sum_{k=1}^{n} a_k$ Prove that $s_n \to a$.

So $(a_n-a)$ is a null sequence. I want to show that $(s_n-a)$ is a null sequence. By a previous exercise, I know that $(x_n)$ is a null sequence $\implies$ $(y_n)$ is a null sequence where $y_{n} = (x_1+ \cdots+ x_n)/n$. So can we do something analogous to "adding $a$ to both sides" to get the desired result?

Show that the sequence $a_n = \left(1- \frac{1}{2} \right) \left(1- \frac{1}{3} \right) \cdots \left(1- \frac{1}{n+1} \right)$ is convergent.

So $a_1 = \frac{1}{2}$, $a_2 = \frac{1}{3}, \dots, a_n = \frac{1}{n+1}$. So I conjecture that $(a_n)$ is a null sequence. In other words, for each $\epsilon >0$, $|a_n| \leq \epsilon$ for all $n>N$. Let $\epsilon = \frac{1}{n}$. Choose $N = n+1$. Then the convergence follows?

Prove that the sequence $a_n = \frac{1}{n+1}+ \frac{1}{n+2} + \dots + \frac{1}{n+n}$ is convergent to a limit $\leq 1$.

So $a_{n} < a_{n+1}$ for all $n$. Then I need to show that it is bounded above by $1$. To show this should I consider $(1-a_n)$? All the terms are $<1$.

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    @amWhy: No worries. In general, tagging with "meta" tags is usually discouraged, as they normally don't give an idea as to what subjects the question could fall into. There are discussions about this on meta (both math.meta and meta.stackoverflow).2011-06-23

2 Answers 2

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You got the first one: If $x_n = a_n - a$ then $y_n = \frac{1}{n} \sum x_n = s_n - a$.

And the second one is a telescoping product and you got that right too.

For the third one: Hint: $\frac{1}{n+5} \lt \frac{1}{n+4}$

(Spoiler)

We have $\frac{1}{n+k} \le \frac{1}{n+1}$ for $k \ge 1$ and thus $a_n \le \frac{n}{n+1} \lt 1$

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    @Damien: In the second problem, you noticed that $a_n=\frac{1}{n+1}$. You could have stopped there. The $\epsilon$-$N$ argument is trivial in this case, you have done it before.2011-06-23
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For the last sequence, you could use the fact that

$ \left(1+\frac{1}{n}\right)^n

Take logarithms of this inequality, and you'll get some nice telescopic bounds for $\frac{1}{n}$.

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    Yes, I guessed so :-)2011-06-23