[College Calc 01-02. Limits of Sequences] #11

Calculus Level 4

In order to prove that $\displaystyle \lim_{n\to\infty}a_n$ diverges, we first suppose that it converges to some limit $L,$ and find an appropriate $\epsilon>0$ such that for every pair of real values of $N$ and $L$ the statement

$n>N\implies |a_n-L|<\epsilon$

is false, which would be a contradiction to the supposition that $a_n$ is convergent.

Choose all valid sequences $\{a_n\}$ so that $\epsilon=1$ would yield such a contradiction.

$\begin{aligned} {\rm A.} &&& a_n = (-1)^n \\ {\rm B.} &&& a_n = \sin n \\ {\rm C.} &&& a_1 = \frac{1}{10},~a_{n+1} = \begin{cases}a_n+\frac{1}{n+\sqrt{n}} & (\sqrt{n}\textrm{ rational}) \\ -a_n & (\textrm{otherwise})\end{cases} \end{aligned}$

This is a part of the College Calc problem set. You can find more problems here .

None of them

\rm A

only

\rm B

only

\rm C

only

\rm A,~B

only

\rm B,~C

only

\rm A,~C

only All of them

1 solution

K T
Dec 11, 2020

A. Since terms will be always alternating between -1 and 1, there is no L that they will become arbitrarily close to. But the question is not asking whether the limit diverges, but whether a choice of $ε=1$ will suffice to show that. Setting $L=0$ we see that it does: $|a_n - 0| = 1$ . Any other choice for L will only make $|a_n-L|$ larger for either even or odd values of n. So case A contradicts the given statement when $ε=1$ .

B. A similar argument. Terms will keep going up and down taking values between -1 and 1 without converging, but the difference with A is, that values will never quite reach the values -1 or 1, because n would have to be equal to $n=(k+\frac{1}{2})π$ for $k \in \mathbb{Z}$ . Since π is irrational this won't happen. We find $|a_n - 0| \lt 1$ . So case B does not contadict the given statement for $ε=1$ .

C. A tricky one, we have to be careful.

First observe that 0 is the best choice for L because there will always be consecutive terms changing sign and 0 has the least maximum value for $|a_n-L|$ .

Next observe that the sign keeps flipping all the time until $n-1$ is a perfect square. (N-1?? Yes, the definition has a special case for $a_{n+1}$ when n is a perfect square - roots of natural numbers are either irrational or natural).

Next observe that the separation between consecutive squares is always an odd number. That means there will be an even number of sign flips and then adding a term, so all the terms are eventually adding up in the positive sense.

Next notice that, although $\frac{1}{10}+\sum_{k=1}^{\infty}\frac{1}{k+k^2}$ exists, we can see that it is bigger than 1. Its partial sum reaches 1 when $k=9$ . This means that terms $a_{n}$ with $n>=82$ will have $|a_n-0|>=1$ . They are still flipping sign, so setting L closer to a specific value will not help. So for any choice of $L$ and $N$ , $|a_n - L| >= 1$ at least for some $n>N$ and case C contradicts the given statement for $ε=1$ .

Summary

All three sequences actually diverge, but for A and C even the course choice of $ε=1$ suffices to show that.

Bonus

Sequence C is divergent, but still 'absolute convergent', meaning that $\exists L: \forall ε>0: \exists N: \forall n>N: ||a_n|-L|<ε$ . Its absolute value converges to this value $L=\frac{11}{10}$

You've written up such a nice solution, thanks! I really appreciate it.

Boi (보이) - 5 months, 3 weeks ago

Thank you. It is a good problem, quite deceptively looking easy, and demonstrating the care that is needed when dealing with limits.

K T - 5 months, 3 weeks ago

[College Calc 01-02. Limits of Sequences] #11

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