5
$\begingroup$

Let $a_1, a_2, \dots$ be a sequence of real numbers such that the sequence $a_1 + 2a_2, a_2 + 2a_3, a_3 + 2a_4 \dots$ converges. Prove that the sequence $a_1, a_2, \dots$ must also converge.

I tried rewriting the $a$ sequence in terms of the first convergent sequence. Letting $$b_1 = a_1 + 2a_2, \ b_2 = a_2 + 2a_3, \ b_3 = a_3 + 2a_4, \dots$$ we have $$a_2 = \frac{b_1 - a_1}{2},$$ $$a_3 = \frac{b_2 - a_2}{2} = \frac{2b_2 - b_1 + a_1}{4}, $$ $$a_4 = \frac{b_3 - a_3}{2} = \frac{4b_3 - 2b_2 + b_1 - a_1}{8}, \\ \dots$$

I have no experience in analysis so I tried using only the definition of convergence and $\lim_{n \to \infty}a_n$. My definition of convergence is for any $\epsilon > 0$ there exists $N$ such that $|b_n - c| < \epsilon$ for $n \ge N$. However even though I can get a magnitude bound on the alternating sum of $b_1, b_2, \dots$, I'm still unsure of how to deal with the finitely many $b_1, \dots, b_{N-1}$ that aren't within $\epsilon$ of $c$. I think Cauchy sequences might help but I have no experience with them either.

$\endgroup$
  • $\begingroup$ This is Putnam 2, number 3 here. So see Putnam solutions. $\endgroup$ – Dietrich Burde Oct 19 '17 at 19:50
  • $\begingroup$ @DietrichBurde As far as I know those aren't real Putnam questions. This one seems too easy anyway. $\endgroup$ – qwr Oct 19 '17 at 19:53
  • $\begingroup$ Can we write $B_n: b_1,b_2,...$ as the weighted sum of two sequences: $A_n: a_1,a_2,..$ and $A'_n: a_2,a_3,....$? Then argue that if $A_n$ is converging to say $C$, then $A'_n$ is converging to the same point $C$. Hence, $A_n+2*A'_n=B_n$ converges to $3C$. Similarly, if $A_n$ is diverging then $B_n$ as well. Now, Can't we just prove the opposite? If $a_1,a_2,...$ diverges then $b_1,b_2,...$ diverges? Then, We have the results. $\endgroup$ – A. M. Oct 19 '17 at 21:06
3
$\begingroup$

Let $\epsilon>0$. Then we find $N$ such that $|b_n-c|<\frac12\epsilon$ for all $n\ge N$. Consider $m>N$ and let $r=\frac{|a_m-\frac13c|}\epsilon$. Then we can show by induction that $$ \frac{\left|a_{m-k}-\frac13c\right|}\epsilon\ge 2^k\left(r-\frac12\right)+\frac12$$ for $0\le k\le m-N$. The essential step here is to note that $$ \left|a_{n-1}-\frac13c\right|=\left|b_{n-1}-c-2\Bigl(a_n-\frac13c\Bigr)\right|\ge 2\left|a_n-\frac13c\right|-\left|b_{n-1}-c\right|.$$

In particular, $$ \frac{\left|a_{N}-\frac13c\right|}\epsilon\ge 2^{m-N}\left(r-\frac12\right)+\frac12.$$ With $\epsilon$ and $N$ fix, $m$ must be bounded if $r>\frac12$. Therefore $r<1$ for all large enough $m$, in other words $a_n\to \frac13c$.

$\endgroup$
  • $\begingroup$ I wonder, how did you arrive at $c/3$ as limit? Though I guess letting $b_1 = b_2 = \dots = 1$ resolves to $c/3$. $\endgroup$ – qwr Oct 22 '17 at 17:36

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.