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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    $\begingroup$ @amWhy: Yes. Should I write something else? $\endgroup$ – Damien Jun 23 '11 at 18:04
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    $\begingroup$ @amWhy: One reason is that Damien actually seems to have shown efforts to solve each problem he has asked so far, the questions are well formatted and he shows genuine interest in learning. Of course, by taking it slower (i.e. giving more thought rather than asking here) would only help Damien and he should consider asking only when he is really really stuck. I agree with that, if that is where you are coming from. $\endgroup$ – Aryabhata Jun 23 '11 at 18:09
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    $\begingroup$ @amWhy: I have no teacher to check my work. Also the book doesnt have solutions. Hence I post some problems here. $\endgroup$ – Damien Jun 23 '11 at 18:10
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    $\begingroup$ Damien: There's nothing wrong about asking questions; that's what this site is for. But I become a little concerned when one user posts as many questions as you have in such short a period of time. Whether self-study or homework...I don't know if its such a good idea to become too "dependent" on others for solutions. To your credit, you do show work, speculate, etc., and we really appreciate that. Sometimes struggling a little while longer with a problem before asking for help can go a long way...and helps you better clarify the concepts that are really problematic for you? $\endgroup$ – Namaste Jun 23 '11 at 18:13
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    $\begingroup$ Yes, @Aryabhata...see my follow up post which I was writing before I saw your comment. I did take note: to Damien's credit, he really has done a good job of showing work, speculating, engaging...etc. To Damien: I understand; it can get really frustrating when you have no where to turn. I'm not suggesting you stop posting questions; not at all. But sometimes working through a set of exercises, first, helps you identify conceptual difficulties, and choosing one or two exemplary exercises to ask about may prevent you from asking about "A?", then having to post a follow up "A II?" $\endgroup$ – Namaste Jun 23 '11 at 18:17

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: $\frac{1}{n+5} \lt \frac{1}{n+4}$


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

  • $\begingroup$ @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. $\endgroup$ – André Nicolas Jun 23 '11 at 18:55

For the last sequence, you could use the fact that

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

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

  • $\begingroup$ Do you mean $\frac{1}{n+1} \lt \log (n+1) - \log n \lt \frac{1}{n}$? One easy way to see that is to apply the mean value theorem... +1 though. $\endgroup$ – Aryabhata Jun 23 '11 at 19:21
  • $\begingroup$ yes, that's it. Nice proof. Mine is just without using differentiability. $\endgroup$ – Beni Bogosel Jun 23 '11 at 19:25
  • $\begingroup$ Yes, I guessed so :-) $\endgroup$ – Aryabhata Jun 23 '11 at 19:30

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