Show that the sequence $a_n=1-\frac13+\frac{1}{3^2}-...+(-1)^n\frac{1}{3^n}$ is bounded. 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 $$S_n=\dfrac{\left(-\frac13\right)^n-1}{-\frac43}$$

Using the expression you obtained,
$$|a_n|=\frac34\left|\frac{(-1)^n}{3^n} -1\right|\leqslant\frac34\left(\frac1{3^n}+1\right)\leqslant1.$$
