Mathematical Analysis, Sequences, Limits, Convergence

Asked Jun 29 '23 at 14:52

Active Jun 29 '23 at 15:36

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Suppose $x_{n}$ is a sequence of nonnegative real numbers satisfying $x_{n+1}\le x_{n}+\frac{1}{n^{2}}$ for all $n\ge 1$. Prove that $\lim_{n\to\infty}x_{n}$ exists.

My idea was that $x_{n+1}-x_{n}<\frac{1}{n^{2}}$ and from this point I can conclude that by telescoping series if exist $\lim_{n\to\infty}x_{n}$ is bounded by using the right side but I could not conclude the existence of limit exactly. I will appreciate if you help. Although this seems easy question I could not find the exact idea becuase $(x_{n+1}-x_{n})$ does not need to be positive.

edited Jun 29 '23 at 15:36

Kroki

13,135

asked Jun 29 '23 at 14:52

maths and chess

Please use more descriptive titles – FShrike Jun 29 '23 at 14:53
Remember that the real numbers form a complete set so Cauchy sequences converge. – John Douma Jun 29 '23 at 14:56
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@JohnDouma Note there is a difficulty in that we don't know what $|x_{n+1}-x_n|$ looks like, only what $x_{n+1}-x_n$ looks like – FShrike Jun 29 '23 at 14:56
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Here is another answer to your question. To avoid posting duplicates, the site https://approach0.xyz/ is very useful. – Mittens Jun 29 '23 at 15:20
@Mittens this question is different as $x_n$ is nonnegative – Tim Jun 29 '23 at 15:20
@Tim: look then at this other posting. – Mittens Jun 29 '23 at 15:22
@Mittens Yes, I think this answer the question. Anyway, it is nice to see FShrike answer. – Tim Jun 29 '23 at 15:27
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It happens too often that I answer a question in good faith and it turns out the OP did not want to check if the question had been already answered first. I've posted my answer to one of the duplicates now @Tim – FShrike Jun 29 '23 at 15:46
I was trying to understand it where did you post could you send me a link? – maths and chess Jun 29 '23 at 15:49

Mathematical Analysis, Sequences, Limits, Convergence

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