I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:

Attempt 1:

Write out the terms of $b_n$:

$$b_1=a_{2}-\frac{a_{1}}{2}$$ $$b_2=a_{3}-\frac{a_{2}}{2}$$ $$b_3=a_{4}-\frac{a_{3}}{2}$$ $$\cdots$$ $$b_n=a_{n+1}-\frac{a_{n}}{2}$$

Adding up the terms you get:

$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$

But a dead end here.

Attempt 2:

For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$

Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges.

For $m>n>K$, \begin{align*} |a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\ &\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\ &\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\ &<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|, \end{align*} and a dead end.


5 Answers 5


Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then

$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$

Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any fixed $m$ and for any $n > m$, we have

$$ |a_n| \leq \Bigg| \underbrace{b_{n-1} + \cdots + \frac{b_{n-m}}{2^{m-1}}}_{\text{(1)}} \Bigg| + \underbrace{\frac{|b_{n-m-1}|}{2^m} + \cdots \frac{|a_1|}{2^{n-1}}}_{(2)}.$$

Note here that

  • $\text{(1)}$ consists of fixed number of terms, each tending to zero as $n\to\infty$.

  • $\text{(2)}$ is uniformly bounded by $\frac{M}{2^m} + \frac{M}{2^{m+1}} + \cdots = \frac{M}{2^{m-1}}$.

So, taking limsup as $n\to\infty$ yields

$$ \limsup_{n\to\infty} |a_n| \leq \frac{M}{2^{m-1}}. $$

Since the LHS is a fixed number and $m$ is arbitrary, letting $m\to\infty$ proves the claim.

  • $\begingroup$ Thanks for the answer. Can you perhaps shed some light on how you got the intuition for expressing the $a_n$ in term of the $b_n$'s?? I tried so much but I didn't succeed but once I saw the expression the proof made sense. Any idea about the intuition involved? $\endgroup$ Apr 27, 2017 at 15:15
  • 1
    $\begingroup$ @AspiringMat, I am pretty sure that you already have the idea in your mind. In your last attempt, you are adding and subtracting terms of the form $\frac{a_{n-k}}{2^k}$ in order to relate $a_n - a_m$ to $b_n$'s. My expression on $a_n$ came from exactly the same idea. $\endgroup$ Apr 27, 2017 at 15:20

Let $c_n = 2^n a_n$. The assumption reads $$ \frac{c_{n+1}-c_n}{2^{n+1}-2^n} \to 0. $$ The sequence $2^n$ tends monotonically to $+\infty$, hence by the Stolz–Cesàro theorem $$ a_n = \frac{c_n}{2^n} \to 0. $$

  • $\begingroup$ Very concise solution. (+1). $\endgroup$ Apr 27, 2017 at 14:18

Let $\epsilon > 0$.

Since $a_{n+1}-a_n/2$ converges to $0$, there is an integer $m$ such that for any $n \ge m$, $|a_{n+1}-a_n/2| \le \epsilon/4$.

For such an $n$, you have
$|a_{n+1}| - \epsilon/2 \\ \le |a_{n+1} - a_n/2| + |a_n/2| - \epsilon/2 \\ \le \epsilon/4 + |a_n|/2 - \epsilon/2 \\ = |a_n|/2 - \epsilon/4 \\ = (|a_n| - \epsilon/2)/2$

Intuitively you can interpret this as something saying that $|a_n|$ has to decrease somewhat exponentially at least until $|a_n|$ gets too close to $\epsilon/2$.

Then let us show there is an $m' \ge m$ such that $|a_{m'}| \le \epsilon$.

If $|a_m| \le \epsilon$ then we are already done by picking $m'=m$, so suppose $|a_m| > \epsilon$.
Now, $(|a_m|- \epsilon/2) / (\epsilon/2) > 1 > 0$ so there is an integer $k$ such that $2^k \ge (|a_m|- \epsilon/2) / (\epsilon/2)$.
Looking at $m' = m+k$ we get
$|a_{m+k}| - \epsilon/2 \le (|a_m| - \epsilon/2) 2^{-k} \le \epsilon/2$,
and so $|a_{m'}| \le \epsilon$.

Then we can prove by induction that for any $n \ge m'$, $|a_n| \le \epsilon$ :

This is true for $n=m'$.
Suppose $n \ge m'$ and $|a_n| \le \epsilon$.
Then $|a_n|-\epsilon/2 \le \epsilon/2$, and so because $n \ge m$,
$|a_{n+1}| - \epsilon/2 \le (|a_n| - \epsilon/2)/2 \le \epsilon/4 < \epsilon/2$, and finally $|a_{n+1}| \le \epsilon$.

Therefore, for all $n \ge m', |a_n| \le \epsilon$, and we have shown that the sequence $a_n$ converges to $0$.


Let $b_n=a_{n+1}-(a_{n}/2)$ so that $b_{n} \to 0$ as $n\to\infty$. Now we have $a_{2}=b_{1}+(a_{1}/2),a_{3}=b_{2}+b_{1}/2+a_{1}/4$ and continuing in this manner we get $$a_{n+1}=b_{n}+b_{n-1}/2+\cdots+b_{1}/2^{n-1}+a_{1}/2^{n}$$ Let $\epsilon>0$ be given then we can see that there is a positive integer $m$ such that $|b_n|<\epsilon$ for all $n\geq m$. Thus if $n>m$ then we have $$|a_{n+1}|\leq |b_{n} |+|b_{n-1}|/2+\cdots+|b_m|/2^{n-m}+|b_{m-1}|/2^{n-m+1}+\cdots +|b_{1}|/2^{n-1}+|a_{1}|/2^{n}$$ First $(n-m+1) $ terms on right are together less than $2\epsilon$ and the rest of $m$ terms tend to $0$. Thus taking $\limsup$ on both sides we see that $\limsup|a_{n+1}|\leq 2\epsilon$ and since $\epsilon$ was arbitrary this means that $a_{n} \to 0$ as $n\to\infty$.


Hint : If $L_t n > \infty$ $\frac{A_{n+1}}{A_{n}}$ is less than $1$. Then the sequence $A_n$ must converge.

  • 2
    $\begingroup$ I am not sure how this helps? $\endgroup$ Apr 27, 2017 at 13:15
  • $\begingroup$ Divide the given equation by an $\endgroup$ Apr 27, 2017 at 13:19
  • $\begingroup$ You can choose $(a_n)$ so that $a_{n+1} - \frac{1}{2}a_n$ converges to $0$ but $a_{n+1}/a_n$ diverges (and even has $+\infty$ as a limit point). Does your hint cover this case? $\endgroup$ Apr 27, 2017 at 14:11
  • $\begingroup$ Its a hint. Its not supposed to provide an answer.Also An+1/An as limit n tends to infinite is equal to 1/2 from the equation. This shows that the later terms of the series are lesser than the initial terms .Thus it proves that the series is converging. $\endgroup$ Apr 28, 2017 at 2:56

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