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I have this exercise from the J.C.Burkill book on Mathematical Analysis I'm struggling to get started on. It is about limit of sequences.

Basically, it says $\lim \dfrac{(s_{n+1})}{s_n} = l;$ $ -1 < l < 1,$

I have to prove that $s_n $ converges to $ 0$.

I would like to know if I can start proving this using the $|s_n - k | < \epsilon$.

My first thought was to start with:

$$\left|\frac{s_n}{s_{n-1}} - l\right| < E$$

Then,

$$-l -E < \left|\frac{s_{n}}{s_{n+1}} \right| < l + E$$

If not, would appreciate if someone could help me get started.

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  • $\begingroup$ wow, just got half of my question deleted. Is it wrong to show my line of thinking around here? I would have thought that by posting how I'm trying to solve it would help someone to show why that is not the right way of thinking. Thanks for the fixing the symbols, though. $\endgroup$ – nightcoder Mar 28 '15 at 11:33
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    $\begingroup$ No it is not. In fact showing your work is appreciated. I suggest you edit your question and add that part back in. I wonder why @learnmore edited that out. $\endgroup$ – Arpan Mar 28 '15 at 11:39
  • $\begingroup$ The question should be stated and edited more carefully; to start with, it should be mentioned that $(s_n)$ is a given sequence of real numbers. More importantly, what is "$(s_{n+1})$" in the third line? Is it just a typo and the brackets $(\ )$have no meaning, or is there something which is not mentioned? $\endgroup$ – user 59363 Mar 28 '15 at 12:04
  • $\begingroup$ I just typed the question as it is printed on the book. I understand that this may be incomplete, although I (perhaps erroneously) would expect that being real analysis and sequences, it would be possible to understand the question even without the details you mentioned. But I agree with you that questions should be as precise as possible. $\endgroup$ – nightcoder Mar 28 '15 at 12:09
  • $\begingroup$ I add to my previous comment: What is "$E$"? Are the "$S_n$" the same as the "$s_n$" or something new? $\endgroup$ – user 59363 Mar 28 '15 at 12:11
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If you know the ratio test for series, the answer is nice: $\sum_n s_n <\infty$ and hence $\lim_n s_n =0$.

Otherwise, just mimic the proof of the ratio test: $$ |s_2| \leq (|\ell| +\epsilon) |s_1|, \quad |s_3|<(|\ell|+\epsilon) |s_2| < (|\ell|+\epsilon)^2 |s_1|,\ldots $$ By induction, $$ |s_{n+1}| < (|\ell|+\epsilon)^n |s_1|, $$ and if you take $\epsilon>0$ so small that $|\ell|+\epsilon<1$ you conclude.

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