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.

  • $\begingroup$ This is Putnam 2, number 3 here. So see Putnam solutions. $\endgroup$ Oct 19, 2017 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, 2017 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, 2017 at 21:06

1 Answer 1


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 $a_n\to \frac13c$.

  • $\begingroup$ 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, 2017 at 17:36

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