How Does the Ratio Test Prove Series Convergence?

[Bipolarity]

I am trying to understand something in the proof of the ratio test for series convergence.
If $a_{n}$ is a sequence of positive numbers, and that the ratio test shows that $\lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r < 1$, then the series converges.

Apparently, the proof defines a number R : r<R<1, and then shows that there exists a N>0 such that $\frac{a_{n+1}}{a_{n}} < R$ for all n>N. It need not to be true in the case where n=N, right? Up to this part I get.

But then it concludes from the above that, there exists a positive N such that
$a_{N+1}<a_{n}R$ which does not follow due to the statement in bold.

Could someone please point out where I am wrong so I can continue this theorem without any qualms? Thanks!

BiP

 

[MarneMath]

Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got.

 

[Bipolarity]

Hmm, think about what it means for the limit to exist. I think that might be the missing piece that shows why it works. Because if you believe it exist, there is an n greater than or equal to N such that (a_(n+1))/(a_n) is less than or equal to R for n greater than N. Then from there you can rewrite the inequality to get what you got.

Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand.

BiP

 

[micromass]

Are you sure the part in bold is correct? Doesn't the limit definition exclusively use greater than? Because that is precisely what I don't fully understand.

BiP

It doesn't really matter. The statement:

For all $\varepsilon>0$, there exists an N such that for all $n\geq N$ holds that $|a_n-a|<\varepsilon$.

is actually equivalent with

For all $\varepsilon>0$, there exists an N such that for all $n> N$ holds that $|a_n-a|<\varepsilon$.

So you can use both statements to define limit of a sequence. Of course, once you decided on which of both versions to use, you have to be consistent.

 

[blue_raver22]

olarBear

I can understand your confusion and I would be happy to help clarify the proof for you. First, let's review the statement of the ratio test: If the sequence of positive numbers {a_n} satisfies \lim_{n→∞}\frac{a_{n+1}}{a_{n}} = r < 1, then the series \sum_{n=1}^{\infty} a_n converges.

In the proof, we start by defining a number R such that r < R < 1. Then, we show that there exists a positive integer N such that for all n > N, we have \frac{a_{n+1}}{a_{n}} < R. This is because the limit of the ratio is r, which is less than R, so eventually the terms in the sequence will be less than R. This is where your understanding is correct.

However, the statement in bold is not a conclusion, but rather a clarification of the previous statement. It is saying that for n > N, the ratio \frac{a_{n+1}}{a_{n}} is less than R, but this does not necessarily hold true for n = N. In other words, we cannot guarantee that the ratio is less than R for the first N terms of the sequence. This is why we specify that this is true "for all n > N".

Finally, we use this information to show that there exists a positive integer N such that a_{N+1} < a_nR. This follows from the fact that for all n > N, we have \frac{a_{n+1}}{a_{n}} < R. Therefore, if we multiply both sides by a_n, we get a_{n+1} < a_nR. This holds true for all n > N, so we can choose N such that this inequality holds for all n > N.

I hope this clarifies the proof for you. Remember, mathematical proofs can be complex and it is important to pay attention to the details and understand each step. Keep practicing and you will become more comfortable with these concepts.