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I must show that the sequence $a_n = \frac{n^2}{4^n}$ is convergent. I'm taking real analysis for the first time and could use some help. Here is what I have so far: Assume $a_n$ converges to $L$ and let $\epsilon > 0$. Then by definition we have$$\left|\frac{n^2}{4^n} - L\right| < \epsilon \to \left|\frac{n^2}{4^n} - \frac{L(4^n)}{4^n}\right| < \epsilon \to \left|\frac{n^2 - L(4^n)}{4^n}\right| < \epsilon.$$I'd like to find some natural number $M$ such that for all $n > M$, the definition holds true. I'm not sure how to proceed. I am kinda lost. I would appreciate any help.

user
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Jake Alge
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  • We can guess that $L=0$ for example. Are you forced to use $\varepsilon-\delta$? – user Sep 16 '21 at 21:06
  • I'm not forced to use that, no. Do I need to show that the limit of the sequence is $0$ before guessing that it is in the proof? – Jake Alge Sep 16 '21 at 21:19
  • When we proceed directly by $\varepsilon-\delta$ method we usually guess the value for $L$. In this case it seems reasonble to take $L=0$. Which theorem are you allowed to use at this stage? – user Sep 16 '21 at 21:23

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If you know the ratio test then you may approach as follows. $$\frac {a_{n+1}}{a_n} = \frac {(n+1)^2}{n^2}* \frac {4^n}{4^{n+1}}$$ $$\lim _{n\to \infty } \frac {a_{n+1}}{a_n} =1/4 <1$$ Thus the series $$ \sum _{n=1}^\infty a_n $$converges which implies $$\lim _{n\to \infty } a_n =0$$ You really do not need the ratio test for series because once we have $$\lim _{n\to \infty } \frac {a_{n+1}}{a_n} =1/4 <1$$ For some natural $m$ we have $$n>m\implies 0< \frac {a_{n+1}}{a_n}<1/2$$ For $n>m$ we get $$|\frac {a_n}{a_m}|<(1/2)^{n-m}$$ Which implies $$\lim _{n\to \infty } a_n =0$$

Mohammad Riazi-Kermani
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    The OP says he/she is taking real analysis for the first time. It's not right to expect him/her to know the ratio test. – Sayan Dutta Sep 16 '21 at 21:21
  • @Sayn I thought calculus comes before real analysis – Mohammad Riazi-Kermani Sep 16 '21 at 21:23
  • It is unclear if OP is talking the series or the sequence converging. Certainly, the $L$ in OPs questions can only be the limit of $a_n.$ – Thomas Andrews Sep 16 '21 at 21:24
  • @mohammad I thought ratio test is taught after completing an introductory course to real analysis – Sayan Dutta Sep 16 '21 at 21:25
  • @SayanDutta Easy fix, with citation to Ratio Test. To Mohammad Riazi-Kermani However, since the OP seems to be questioning the convergence of a sequence rather than a series, I am wondering if you could simplify your answer to (alternatively) merely demonstrate from scratch that $a_n \to 0$. – user2661923 Sep 16 '21 at 21:26
  • @user2661923 I felt, the OP needed an $\epsilon-\delta$ proof. But, well, citation also works I guess :| – Sayan Dutta Sep 16 '21 at 21:27
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It can be shown by induction (or calculus) that $n^2\leq 2^n$ for $n\geq 4$. Note that $$0\leq \left|\frac{n^2}{4^n}\right|=\frac{n^2}{4^n}\leq \frac{2^n}{4^n}=\cdots$$

C Squared
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Use Induction to show that $$4^n>n^3\;\;\forall n\in \mathbb N$$ So, $$\frac {n^2}{4^n}<\frac 1n$$ Also, clearly, $$\frac 1{4^n}<\frac {n^2}{4^n}$$

Now, use the Sandwich Theorem (or Squeeze Theorem or whatever you call it) to arrive at the answer. I assumed that you know about the convergence of the two sequences $\left\{\frac 1n\right\}$ and $\left\{\frac 1{4^n}\right\}$ (to $0$)

Sayan Dutta
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