3
$\begingroup$

I have a sequence $a_{n}=\frac{(n+1)(n^{2})}{(2n+1)(3n^{2}+1)}$ ,

and limit of it when $n$ goes to infinity is $\frac{1}{6}$.

Because limit is some number, is that enough of a proof that this sequence is convergent, or should i do something more?

$\endgroup$
  • 1
    $\begingroup$ It is enough. your sequence CONVERGES to $\frac{1}{6}$. $\endgroup$ – hamam_Abdallah Oct 23 '16 at 14:10
  • 3
    $\begingroup$ This is the definition of convergence, so what you say is ok. $\endgroup$ – zar Oct 23 '16 at 14:10
  • $\begingroup$ You could also argue that $a_n$ can be written as the quotient of two convergent sequences whose denominator has nonzero limit, hence $a_n$ is convergent. $\endgroup$ – lzralbu Oct 23 '16 at 14:18
3
$\begingroup$

It is enough to write, for $n>1$, $$ a_{n}=\frac{(n+1)(n^{2})}{(2n+1)(3n^{2}+1)}=\frac16\cdot\frac{1+\frac1n}{\left(1+\frac1{2n}\right)\left(1+\frac1{3n^2}\right)} $$ giving, as $n \to \infty$, $$ a_n \to \frac16\cdot\frac{1+0}{\left(1+0\right)\left(1+0\right)}=\frac16. $$

$\endgroup$
  • $\begingroup$ Thank you, i needed conformation because im confused when i need to use that $\epsilon$ proof and when determining the limit is enough. $\endgroup$ – Tars Nolan Oct 23 '16 at 14:14
  • $\begingroup$ @TarsNolan You are welcome. $\endgroup$ – Olivier Oloa Oct 23 '16 at 14:15
  • $\begingroup$ But isn't there a theorem that says that if u(n) tends to 0 as n tends to infinity the series can be convergent but not necessary and that's why we use the ratio test and root test. But if u(n) doesn't tend to 0 as n tends to infinty then sure shot the series is divergent. Am I wrong because even I had this doubt ? $\endgroup$ – Shashaank Oct 23 '16 at 14:21
  • 1
    $\begingroup$ @Shashaank I think you are making a confusion between convergence of a sequence $\left\{u_n\right\}_{n\ge 0}$ and convergence of a series $\sum_{n\ge 0} u_n$. $\endgroup$ – Olivier Oloa Oct 23 '16 at 14:39
  • $\begingroup$ @OlivierOloa Ok so is it that what I am saying is about the convergence of a series and the question and the answers addresse the convergence of a sequence ? $\endgroup$ – Shashaank Oct 23 '16 at 14:47
4
$\begingroup$

Theorems on limits lift up the burden of doing $\varepsilon$-$\delta$ proofs. If you know that $\lim_{n\to\infty}a_n=l$ and $\lim_{n\to\infty}b_n=m$ (real $l$ and $m$), then you can also say \begin{gather} \lim_{n\to\infty}(a_n+b_n)=l+m \tag{Theorem 1}\\[6px] \lim_{n\to\infty}(a_nb_n)=lm \tag{Theorem 2}\\[6px] \lim_{n\to\infty}\frac{a_n}{b_n}=\frac{l}{m} \tag{Theorem 3} \end{gather} Theorem 3 requires the hypothesis $m\ne0$ (so also $b_n\ne0$ from some point on).

Such theorems can also be extended to the cases when one or both the limits are infinity, but this would take too far.

In your case, the sequence $a_n$ is not, as written, in one of the above forms, but you can rewrite it as $$ a_n=\frac{n+1}{2n+1}\frac{n^2}{3n^2+1} $$ Now let's examine $$ b_n=\frac{n+1}{2n+1} $$ We can't apply the third theorem above, but as soon as we rewrite $$ b_n=\frac{1+\frac{1}{n}}{2+\frac{1}{n}} $$ we see that at numerator and denominator we have sequences to which we can apply the first theorem above, because we know that $\lim_{n\to\infty}\frac{1}{n}=0$. Thus, combining theorems 1 and 3, we get $$ \lim_{n\to\infty}b_n=\frac{1}{2} $$ Similarly $$ \lim_{n\to\infty}\frac{n^2}{3n^2+1}= \lim_{n\to\infty}\frac{1}{3+\frac{1}{n^2}}=\frac{1}{3} $$ Now apply theorem 2 and the requested limit is $\frac{1}{6}$.

More simply, you can directly do as in Olivier Oloa's answer. Mentioning the application of the above theorems is usually omitted (like Olivier did).

You don't need to check the limit with an $\varepsilon$-$\delta$ proof, because you're applying theorems that have been proved correct and the known fact that $\lim_{n\to\infty}\frac{1}{n}=0$.

$\endgroup$
2
$\begingroup$

I will just elaborate on Olivier Oloa's answer. In the comments you mention you are unsure whether you should use "that $\epsilon$ proof". You don't have to (but sort of are anyway) and here's why.

The definition of $\lim_{n\to \infty} a_n = a$ is something like this $$\forall \epsilon>0 \;\exists n_0\in \mathbb N,\; \forall n \geq n_0: |a_n - a| < \epsilon$$

Now, if you had just that and nothing else, you would indeed need to prove that your sequences converges to $1/6$ using the definition, i.e. "that $\epsilon$ proof"

Luckily, you've probably been shown (or even have proved) some basic results, namely things such as $$\lim_{n\to \infty} \frac 1n = 0$$ and (for $\lim a_n = a$ and $\lim b_n = b$) $$\lim_{n \to \infty} a_n + b_n = a+b$$ and so on.

So altogether what you're doing in the answer you've given is repeatedly using all these already proven theorems and rules to prove what your limit goes to (I recommend very carefully going through your argument to see what exact rules you've used). The theorems themselves are proven by the whole $\epsilon$ machinery, giving you the power to prove the convergence of the sequence without using a single $\epsilon$

Example: Prove that $$\lim_{n\to\infty} \frac 2n = 0$$ Now, you could do this (and quite easily) through the definition, i.e. doing the $\epsilon$ proof. But, using the two facts I wrote down earlier, you could argue $$\lim_{n\to\infty} \frac 2n = \lim_{n\to\infty} \frac 1n + \frac 1n = \lim_{n\to\infty} \frac 1n + \lim_{n\to\infty} \frac 1n = 0 + 0 = 0$$ While this might be a quite convoluted usage (what rule could you use to prove this perhaps more directly?), it illustrates how one escapes the need to prove things from definition by using already proven facts/theorems.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.