En efecto:
\[
\begin{aligned}
n = 1
&\Rightarrow
(-1)^n \frac{2n + 1}{2n – 1}
=
(-1)^1 \frac{ 2(1) + 1 }{2(1) -1}
\\[1em]
&\hspace{8em}=
-3
\\[2em]
n = 2
&\Rightarrow
(-1)^n \frac{2n +1}{2n – 1}
=
(-1)^2 \frac{ 2(2) + 1 }{2(2) – 1}
\\[1em]
&\hspace{8em}=
\frac{5}{3}
\\[2em]
n = 3
&\Rightarrow
(-1)^n \frac{2n + 1}{2n – 1}
=
(-1)^3 \frac{2(3) + 1}{2(3) – 1}
\\[1em]
&\hspace{8em}=
-\frac{7}{5}
\\[1em]
&\hspace{8em}=
\frac{-7}{5}
\\[2em]
n = 4
&\Rightarrow
(-1)^n \frac{ 2m + 1 }{2n – 1}
=
(-1)^4 \frac{2(4) + 1}{2(4) – 1}
\\[1em]
&\hspace{8em}=
\frac{9}{7}
\\[2em]
n = 5
&\Rightarrow
(-1)^n \frac{2n + 1}{2n – 1}
=
(-1)^5 \frac{2(5) + 1}{2(5) – 1}
\\[1em]
&\hspace{8em}=
-\frac{11}{9}
=
-\boldsymbol{ \frac{11}{9} }
\end{aligned}
\]
\[
\begin{aligned}
n = 1
&\Rightarrow
(-1)^n \frac{2n + 1}{2n – 1}
=
(-1)^1 \frac{ 2(1) + 1 }{2(1) -1}
\\[1em]
&\hspace{8em}=
-3
\\[2em]
n = 2
&\Rightarrow
(-1)^n \frac{2n +1}{2n – 1}
=
(-1)^2 \frac{ 2(2) + 1 }{2(2) – 1}
\\[1em]
&\hspace{8em}=
\frac{5}{3}
\\[2em]
n = 3
&\Rightarrow
(-1)^n \frac{2n + 1}{2n – 1}
=
(-1)^3 \frac{2(3) + 1}{2(3) – 1}
\\[1em]
&\hspace{8em}=
-\frac{7}{5}
\\[1em]
&\hspace{8em}=
\frac{-7}{5}
\\[2em]
n = 4
&\Rightarrow
(-1)^n \frac{ 2m + 1 }{2n – 1}
=
(-1)^4 \frac{2(4) + 1}{2(4) – 1}
\\[1em]
&\hspace{8em}=
\frac{9}{7}
\\[2em]
n = 5
&\Rightarrow
(-1)^n \frac{2n + 1}{2n – 1}
=
(-1)^5 \frac{2(5) + 1}{2(5) – 1}
\\[1em]
&\hspace{8em}=
-\frac{11}{9}
=
-\boldsymbol{ \frac{11}{9} }
\end{aligned}
\]