Showing Sequences Converge/Diverge

[ZCohen]

I have had some introduction to set theory and have gone through calculus in a theoretical manner up through first and second order differential equations. However, we are now working on sequences (and series, but I find series to be less of a problem). There doesn't seem to be an easy way to go about showing whether a sequence is convergent or divergent other than using comparison theorems or actually evaluating a limit.

(comparison theorems i.e. x^b = little-o (e^ax) ; a,b > 0)

Is there any way that I am overlooking to evaluate the convergence of sequences?
My question arose when considering the problem

{f(n)} = n² / (n+1) - (n² + 1) / n ,

which I thought to converge to 0 while it actually converges to -1.

 

[alyafey22]

\(\displaystyle f(n)=\frac{n^2}{n+1}-\frac{n^2+1}{n}= \frac{-n^2-n-1}{n^2+n} \)

Then what is the limit of the expression ?

 

[ZCohen]

\(\displaystyle f(n)=\frac{n^2}{n+1}-\frac{n^2+1}{n}= \frac{-n^2-n-1}{n^2+n} \)

Then what is the limit of the expression ?

So this leads me to ask, is a limit the only method of determining convergence of a sequence? And if so, why do some books (in my experience) ask first if a sequence is convergent then ask for the evaluation of the limit?

 

[alyafey22]

So this leads me to ask, is a limit the only method of determining convergence of a sequence? And if so, why do some books (in my experience) ask first if a sequence is convergent then ask for the evaluation of the limit?

If the limit of a sequence exits , then it is convergent.

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In real analysis , you might be asked prove that the sequence is convergent then you can use the definition without needing to find the limit.

 

[Fernando Revilla]

So this leads me to ask, is a limit the only method of determining convergence of a sequence? And if so, why do some books (in my experience) ask first if a sequence is convergent then ask for the evaluation of the limit?

We need to know that technique. Imagine you have the sequence $a_{n+1}=\sqrt{6+a_n},\;a_1=0$ and we are not interested in an explicit formula for $a_n$ (or we can't find it). If the sequence has a limit $L$, then, $L=\sqrt{6+L}$ and solving the equation we get $L=3$ or $L=-2$, so necessarily $L=3$ because $a_n\geq 0$ for all $n$.

We have not yet proved that $a_n$ is convergent. If we prove (for example) that the sequence is increasing and bounded above, then we can assure that the limit is $L=3$.

Also, in Complex Analysis we sometimes need to kow if an improper real integral is previously convergent because in that case, is equal to its Cauchy principal value ... etc. (Perhaps you'll cover this in the future).