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Show that the sequence ${a_n}$ where $$a_n=1-\dfrac13+\dfrac{1}{3^2}-...+(-1)^n\dfrac{1}{3^n}$$ is bounded.

The first thing that came to my mind was to see if the sequence is monotone. If I am right, the $(n+1)th$ term should look like this: $$a_{n+1}=1-\dfrac13+\dfrac{1}{3^2}-...+(-1)^{n+1}\dfrac{1}{3^{n+1}}$$ and then the difference $a_{n+1}-a_n$ is $$a_{n+1}-a_n=\dfrac{(-1)^{n+1}}{3^{n+1}},$$ the sign of which depends on $n$. If we write the first terms $$a_1=1;a_2=\dfrac{2}{3}=\dfrac{6}{9},a_3=\dfrac{7}{9},$$ we can see that the sequence isn't monotone.

Another thing that I noted is that we actually have the sum of a geometric sequence with first term $b_1=1$ and common ratio $-\dfrac13$. That is for the general term, the sum is $$S_n=\dfrac{\left(-\frac13\right)^n-1}{-\frac43}$$

Yajat Shamji
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kormoran
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3 Answers3

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You have shown that $$ a_n = \frac{\left(-\frac13\right)^n-1}{-\frac43} $$ (Note the $a_n$ here, not $S_n$; the $a_n$ themselves are the partial sums of a geometric series.) It is not difficult to see that $\left|\left(-\frac13\right)^n\right|\leq 1$, and together with the triangle inequality we get $$ |a_n| = \left|\frac{\left(-\frac13\right)^n-1}{-\frac43}\right| \leq \frac{\left|\left(-\frac13\right)^n\right| + |-1|}{\left|-\frac43\right|}\leq \frac{1 + 1}{\frac43} = \frac32 $$ which clearly means it's bounded.

Alternately, we can use the monotonicity of every other term. Let $b_n = a_{2n}$ and $c_n = a_{2n+1}$ be the two resulting subsequences. Then $$ b_{n+1} - b_n = a_{2n + 2} - a_{2n} = (-1)^{2n+2}\frac{1}{3^{2n+2}} + (-1)^{2n+1}\frac1{3^{2n+1}}\\ = \frac{1}{3^{2n+1}}\left(\frac13 - 1\right) < 0 $$ so $b_n$ is monotonically decreasing. Similarly we get that $c_n$ is monotonically increasing.

Finally, note that for any $n$ we have $c_n < b_n$. Which by the above monotonicity means that we must have $$ c_0 \leq c_n < b_n \leq b_0 $$ Since every $a_m$ is either a $c_n$ or a $b_n$ for some $n$, we have $c_0 < a_m < b_0$, meaning the original sequence is bounded.

Arthur
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  • Thank you, I think I got it. Can you clarify for me why $\left|\left(-\frac13\right)^n\right|=\left|(-1)^n\left(\frac13\right)^n\right|\le1$? Also $a_n=S_n$ is correct, right? As $S_2$ of the geometric sequence is the sum of the first two terms which is actually the second term of the series... – kormoran Jan 27 '23 at 11:25
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    @KaloyanK. Yes, $a_n = S_n$. And do you mean to tell me that you don't already know that $\frac1{3^n}\leq 1$ is true for all natural $n$? That should've been an exercise long before this problem. – Arthur Jan 27 '23 at 11:29
  • I got it, thank you! I appreciate your help! – kormoran Jan 27 '23 at 11:29
  • @KaloyanK. Note that once we are using absolute values, overall signs don't matter, so the $(-1)^n$ can just be removed. Technically we are using $|(-1)^n\frac1{3^n}| = |(-1)^n|\cdot |\frac1{3^n}|$, and then using that $(-1)^n$ is either $1$ or $-1$, so no matter which one it is, the absolute value is $1$. giving us $|(-1)^n|\cdot |\frac1{3^n}| = |\frac1{3^n}|$. – Arthur Jan 27 '23 at 11:32
  • Actually when $n$ is natural $\left|\dfrac{1}{3^n}\right|<1$. It's equal to $1$ when $n=0$ which isn't natural. – kormoran Jan 27 '23 at 12:23
  • @KaloyanK. Saying that something is $\leq$ when it actually is $<$ is entirely correct. Also, $0$ is a natural number from time to time. It varies from person to person and context to context. – Arthur Jan 27 '23 at 13:08
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Using the expression you obtained,

$$|a_n|=\frac34\left|\frac{(-1)^n}{3^n} -1\right|\leqslant\frac34\left(\frac1{3^n}+1\right)\leqslant1.$$

Angelo
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PierreCarre
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Notice that $$|a_n|<\sum\limits_{k= 0}^{\infty} \frac{1}{3^k}=\frac{3}{2} \ \hspace{0.1cm} \forall n\ge 0.$$

A lot of sequence and series problems can be solved by comparing them to sequence and series that we already know a lot about. This is done rigorously using the comparison tests and squeeze theorems.

Let us look at a few terms here. $$a_1=1\le 1$$ $$a_2=1-\frac1 3<1+\frac1 3<1+\frac1 3 +(\text{Anything positive})$$ $$a_3=1-\frac1 3 +\frac1 {3^2}<1+\frac1 3+\frac1 {3^2}<1+\frac1 3+\frac1 {3^2}+\text{(Anything positive)}$$ and so on.

For a fixed $n$, we have that $$a_n<\sum\limits_{k=0}^{n} \frac1 {3^k}+(\text{Anything positive}).$$ Therefore, we can choose the 'anything positive' carefully to be $$\text{(Anything positive)}=\sum\limits_{k>n} \frac{1}{3^k}.$$ Since the terms are all positive, this is doable.

We can therefore see that each term $$a_n<\sum\limits_{k=0}^{n} \frac1 {3^k}+\sum\limits_{k=n+1}^{\infty} \frac1 {3^k}=\sum\limits_{k=0}^{\infty} \frac1 {3^k}=\frac3 2$$

However, this is merely one direction of the proof. We have only produced an upper bound for the series. We need to produce the lower bound. I will leave that to you.

Umesh Shankar
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