# If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. [duplicate]

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:

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

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.

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.

• 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? – AspiringMat Apr 27 '17 at 15:15
• @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. – Sangchul Lee Apr 27 '17 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 $$a_n = \frac{c_n}{2^n} \to 0.$$

• Very concise solution. (+1). – Sangchul Lee Apr 27 '17 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 $n\to\infty$.

I think we can use basic limit rules:

$\lim_{n \rightarrow \infty}a_{n+1}-\frac{a_n}{2}$ $= \lim_{n \rightarrow \infty}a_{n+1}-\lim_{n \rightarrow \infty}\frac{a_n}{2}$ $= \lim_{n \rightarrow \infty}a_{n+1}-\frac{\lim_{n \rightarrow \infty}a_n}{2} = 0$

To use this, we must show that the sequence converges. I have a failed attempt underneath:

If we assume $a_n$ diverges:

$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} -\frac{1}{2}= \lim_{n \rightarrow \infty}\frac{a_{n+1}-\frac{a_n}{2}}{a_n} \rightarrow \frac{0}{\infty} = 0$

This contradicts our assumption due to the ratio test, and it must thus be false.

• But I think you need to show the sequence converges before you apply the limit rules. The point here should be showing that the sequence converge rather than simply apply the limit rules. – Oscar LIU Apr 27 '17 at 13:26
• Yeah, you are right. So, I have added an attempt at showing convergence. – Avatrin Apr 27 '17 at 13:54
• Indeed, I don't think we are allowed to put a_n in the denominator because we may have a_n being zero for many terms in the tail repeatedly (but surely not converging to 0 for the nonzero terms in the tail). And divergence may not lead to a limit of fixed number (or even infinity). – Oscar LIU Apr 28 '17 at 5:05
• My attempt may be wrong, but I am quite certain it's not for the reason you stated since we are taking the limit. The first few terms do not matter (just like the limit of 1/x as x goes to infinity is zero despite the function not being defined at x=0). So, I am starting a new proof verification question. – Avatrin Apr 28 '17 at 8:23
• To say it more clearly, consider the subsequence of a_n by deleting the zero terms: if the subsequence tends to zero, then we are done. Is that what you mean? In my previous statement, I really meant we have infinitely many but not FEW terms. But later I found that this could be solved in another way (consider the nonzero subsequence). – Oscar LIU Apr 28 '17 at 8:38

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

• I am not sure how this helps? – AspiringMat Apr 27 '17 at 13:15
• Divide the given equation by an – Lakshya Gupta Apr 27 '17 at 13:19
• 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? – Sangchul Lee Apr 27 '17 at 14:11
• 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. – Lakshya Gupta Apr 28 '17 at 2:56