-1
$\begingroup$

enter image description here

We've just started learning sequences and I was faced with this question. I have a strong feeling that I'm missing something and that I'm over simplifying things as these questions almost seem trivial to the point I'm not sure how to correctly prove them..

A. True

Assume $(a_n)_{n=1}^{\infty }$ converges to some $L∈R$.

Then as $\{a_n | n∈N\}$ = $\{b_n | n∈N\}$ for every $n∈N$

$(a_n)_{n=1}^{\infty }=(b_n)_{n=1}^{\infty }=L$

Similarly, if $(b_n)_{n=1}^{\infty }$ converges

$(b_n)_{n=1}^{\infty }=(a_n)_{n=1}^{\infty }=L$.

B. True

Assume $(a_n)_{n=1}^{\infty }$ converges to some $L∈R$.

As $\{n∈N | a_n ≠ b_n\}$ is bounded from above, there exists $M∈N$ such that for every $n≥M$ we have

$a_n = b_n$

Therefore $(a_n)_{n=1}^{\infty }=(b_n)_{n=1}^{\infty }=L$.

Similarly if $(b_n)_{n=1}^{\infty }$ converges, there exists $M∈N$ such that for every $n≥M$ we have

$b_n = a_n$

Therefore $(b_n)_{n=1}^{\infty }=(a_n)_{n=1}^{\infty }=L$.

C. False.

Counter example:

$\{a_n\} = \frac{1}{n}$ for every $n∈N$

$\{b_n\} = Z^{+}$

We've seen in recitation that $(a_n)_{n=1}^{\infty } = 0$. But $(b_n)_{n=1}^{\infty }=∞$, and therefore diverges.

$\endgroup$
3
$\begingroup$

$A)$ is false : take $a_1=-1$, and $a_n=1$ for all $n\geq 2$ and $b_n=(-1)^n$ for all $n\geq 1$ for a counter example.

$B)$ is indeed true, but your justification is strange (what means $(a_n)=(b_n)=L$?) Indeed, $a_n=b_n$ for all $n\geq M$ for some $M\in \mathbb N$. From here, obviously $(a_n)$ converges if and only if $(b_n)$ converges.

$C)$ is true (I even don't understand your counter-example). If $(a_n)$ converges, since $(b_n)$ is increasing, then $(b_n)$ converges as well. If $(b_n)$ converges, then, by squeeze theorem, $(a_n)$ converges as well.

$\endgroup$
2
  • 1
    $\begingroup$ Nice catch, I misread it. $\endgroup$ – user2661923 Mar 13 at 16:50
  • $\begingroup$ Thanks! I was confused about the notation of (A) and as I was about to ask you how your example works it hit me. :) $\endgroup$ – MathCurious Mar 13 at 19:57

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.