Problem source

If $\left|r\right|\neq1,$ show that 
\[
1+r^{2}+r^{4}+\cdots+r^{2n}=\frac{1-r^{2n+2}}{1-r^{2}}\,.
\]
If $r\neq1,$ find an expression for $\mathrm{S}_{n}(r),$ where 
\[
\mathrm{S}_{n}(r)=r+r^{2}+r^{4}+r^{5}+r^{7}+r^{8}+r^{10}+\cdots+r^{3n-1}.
\]
Show that, if $\left|r\right|<1,$ then, as $n\rightarrow\infty,$
\[
\mathrm{S}_{n}(r)\rightarrow\frac{1}{1-r}-\frac{1}{1-r^{3}}\,.
\]
If $\left|r\right|\neq1,$ find an expression for $\mathrm{T}_{n}(r),$
where 
\[
\mathrm{T}_{n}(r)=1+r^{2}+r^{3}+r^{4}+r^{6}+r^{8}+r^{9}+r^{10}+r^{12}+r^{14}+r^{15}+r^{16}+\cdots+r^{6n}.
\]
If $\left|r\right|<1,$ find the limit of $\mathrm{T}_{n}(r)$ as
$n\rightarrow\infty.$

What happens to $\mathrm{T}_{n}(r)$ as $n\rightarrow\infty$ in the
three cases $r>1,r=1$ and $r=-1$? In each case give reasons for
your answer.

Solution source

\begin{align*}
&& S &= 1 + r^2 + r^4 + \cdots + r^{2n} \\
&& r^2S &= \quad \,\,\,\, r^2 + r^4 + \cdots+r^{2n}+r^{2n+2} \\
\Rightarrow && (1-r^2)S &= 1 - r^{2n+2} \\
\Rightarrow && S &= \frac{1-r^{2n+2}}{1-r^2}
\end{align*}

\begin{align*}
&& S_n(r) &= r + r^2 + r^4 + r^5 + r^7 + \cdots + r^{3n-1} \\
&&&= 1 + r + r^2 + \cdots + r^{3n} - (1 + r^3 + r^6 + r^{3n}) \\
&&&= \frac{1-r^{3n+1}}{1-r} - \frac{1-r^{3n+3}}{1-r^3} \\
\\
\Rightarrow && \lim_{n \to \infty} S_n(r) &= \frac{1-0}{1-r} - \frac{1-0}{1-r^3} = \frac{1}{1-r} - \frac{1}{1-r^3} 
\end{align*}

\begin{align*}
&& T_n(r) &= 1 + r^2 + r^3 + r^4 + r^6 + \cdots + r^{6n} \\
&&&= \frac{1-r^{6n+6}}{1-r^6} + \frac{r^2-r^{6n+2}}{1-r^6} + \frac{r^3-r^{6n+3}}{1-r^6} + \frac{r^4-r^{6n+4}}{1-r^6} \\
&&&= \frac{1+r^2+r^3+r^4-r^{6n}(r^2+r^3+r^4+r^6))}{1-r^6} \\
\\
&&\lim_{n \to \infty} T_n(r) &= \frac{1+r^2+r^3+r^4}{1-r^6}
\end{align*}

If $r > 1$ clear it diverges. if $r = 1$ same story. if $r = -1$ the sums in blocks of $4$ are all $1+1-1+1 = 2 > 0$ and so it also diverges.