Chapter 9 - Infinite Series - 9.1 Sequences - Exercises Set 9.1 - Page 607: 52

$|r|<1\Rightarrow$ converges $|r|>1\Rightarrow$ diverges $r=1\Rightarrow$ converges $r=-1\Rightarrow$ diverges

Step 1 1 of 5 We will separate the task into four cases as indicated and determine convergence for each one of them. Step 2 2 of 5 First case: $|r|<1$ The sequence ${r_n}$ is known as a geometric sequence because we have that the ratio between two consecutive terms is always the same, in this case it is $r$. The condition for the convergence of such sequences is that the absolute value of the ratio is less than 1. Since in this case that is true, we have that this sequence converges for $|r|<1$. Moreover, we have that the limit is $\displaystyle\lim_{n\to\infty} r^n = 0,$ $|r|<1.$ Step 3 3 of 5 Second case: $|r|>1$ According to the discussion in the previous step, we have that the geometric sequence converges if and only if the absolute value of the ratio is less than 1. Since that is not the case here, we have that in this case the sequence diverges. Step 4 4 of 5 Third case: $r=1$ In this particular case, we have that all the terms are $1$'s and the sequence has constant terms. Really, we have that $a_n=r^n=1,$ $\forall n\in\mathbb{N}$ and so, we have this $\displaystyle\lim_{n\to\infty} a_n=\lim_{n\to\infty} r^n=\lim_{n\to\infty} 1=1.$ Therefore, the sequence converges to $1$ in this case. Step 5 5 of 5 Fourth case: $r=-1$ In this case we have that the terms will alternate in sign because the subsequence of odd-numbered terms will only contain $-1$ as a term and the subsequence with even-numbered terms will have just $1$'s. Really, for $n=2k$ we have that $a_n=a_{2k}=(-1)^{2k}=1,$ whereas for $n=2k-1$ we have this $a_n=a_{2k-1}=(-1)^{2k-1}=-1.$ Based on this, we have that $\displaystyle\lim_{k\to\infty} a_{2k}=\lim_{k\to\infty} (-1)^{2k}=\lim_{k\to\infty} 1=1$ and $\displaystyle\lim_{k\to\infty} a_{2k-1}=\displaystyle\lim_{k\to\infty} (-1)^{2k-1}=\lim_{k\to\infty} (-1)=-1.$ Finally, since those two subsequences approach different values, according to a theorem, we have that the sequence diverges.