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I recently began real analysis and I just wanted to make sure my proofs make sense.

I know how you prove $ \frac{1}{n}$ but not sure how the constant in the sequence changing from 1 to 5 affects the proof. Below is my take on this:

Prove $ \frac{5}{n+1}$ converges to $0$ as $n$ approaches infinity.

Proof: Given $ϵ > 0$, let $N = \lceil \frac{5}{ϵ}\rceil$ so $\frac{5}{N} ≤ ϵ$

For all $n > N$, we have $$\frac{5}{n+1} ≤ \frac{5}{n} < \frac{5}{N} ≤ ϵ$$

Proof done?

Sebastiano
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Bill Cogn
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  • The information concerning the [tag:solution-verification] tag states “For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof that is in doubt, and why so. This should not be the only tag for a question, and should not be used to circumvent site policies regarding duplicate questions. ” Therefore, your question is off-topic and you should have added at least one other tag. – José Carlos Santos Nov 02 '23 at 10:59
  • @JoséCarlosSantos thank you I added that the constant changing from 1 to 5 is the thing confusing me – Bill Cogn Nov 02 '23 at 11:01
  • As you see, it does not affect it that much. You have proved what you wanted to prove. given $\epsilon > 0$, you found an $N$, the ceiling of $5/\epsilon$, such that for all $n \geq N$ you have $|\frac{5}{n+1}| < \epsilon$. – Neckverse Herdman Nov 02 '23 at 11:05
  • @NeckverseHerdman Great thank you. From my point of view any change is scary, especially when writing formal proofs but thank you – Bill Cogn Nov 02 '23 at 11:06
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    Oops, I hit enter too soon! Yes. In the beginning, writing these proofs is daunting. Making a lot of exercises helps and it will continue to help you as you learn more abstract concepts later on. – Neckverse Herdman Nov 02 '23 at 11:08
  • @NeckverseHerdman May I ask, what is the progression from real analysis? Is it real analysis -> complex analysis -> ? Is there anything in between? – Bill Cogn Nov 02 '23 at 11:13
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    @BillCogn not really, no; there's nothing in between real analysis and complex analysis. It goes Real analysis -> Complex analysis -> Functional analysis -> harder analysis (like non-standard analysis). At least, that's the most common route if you want to go down the analysis path. But basically every mathematician takes real and complex analysis, but that doesn't mean they continue going down the analysis path. – Adam Rubinson Nov 02 '23 at 12:10

1 Answers1

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Firstly we need to write down the formal definition of convergence of sequence. $\{a_n\}$ tends to $L$ if for every $\epsilon >0$, there is a $N$ (that depends on $\epsilon$) such that for all $n \ge N$ we have $|a_n-L|<\epsilon$.

In our case, we need to show $\frac{5}{n+1}$ tends to $0$. We can just try to show (modify the query) that $\frac{5}{n}$ tends to $0$, and then: as (for all $n$) $0 \le \frac{5}{n+1} \le \frac{5}{n}$ - we note this finishes the proof.

Now to show (when $\epsilon >0$ is fixed) that $\frac{5}{n} < \epsilon$, note we can fix $M$ - any positive integer greater than $5/\epsilon$. Then for all $n \ge M$, we see $|\frac{5}{n}|<\epsilon$, hence the proof is finished.

Thanks.

Surajit131
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