Sección 2.3 Sumatoria y Productoria
¶
Definición 2.3.1
Sea \(\{a_{n}\}_{n\in\mathbb{N}},r\leq s\) una sucesión y \(r,s\in\mathbb{N}\text{,}\) entonces definimos a:
i) Sumatoria
\begin{equation*}
\overset{s}{\underset{i=r}{\sum}}a_{i}=a_{r}+a_{r+1}+a_{r+2}+\cdot\cdot
\cdot+a_{s}.
\end{equation*}
ii) Productoria
\begin{equation*}
\overset{s}{\underset{i=r}{\prod}}a_{i}=a_{r}\cdot a_{r+1}\cdot a_{r+2}
\cdot\cdot\cdot a_{s}.
\end{equation*}
Ejemplo 2.3.2
Calcular
1) \(\overset{5}{\underset{i=3}{\sum}}i=3+4+5=12\text{.}\)
2) \(\overset{4}{\underset{i=2}{\sum}}2^{i}=2^{2}+2^{3}+2^{4}=4+8+16=28\text{.}\)
3) \(\overset{6}{\underset{n=3}{\sum}}(n\cdot i)=i\overset{6}{\underset
{n=3}{\sum}}n=i(3+4+5+6)=i(18)=18i\text{.}\)
4) \(\overset{4}{\underset{i=2}{\sum}}\left( \overset{3}{\underset{j=2}{\sum}
}j^{i}\right) \text{,}\) donde
\begin{equation*}
\begin{array}{rl}
\overset{4}{\underset{i=2}{\sum}}\left( \overset{3}{\underset{j=2}{\sum}
}j^{i}\right) \amp =\overset{4}{\underset{i=2}{\sum}}(2^{i}+3^{i})\\
\amp =\overset{4}{\underset{i=2}{\sum}}2^{i}+\overset{4}{\underset{i=2}{\sum}
}3^{i}\\
\amp =(2^{2}+2^{3}+2^{4})+(3^{2}+3^{3}+3^{4})\\
\amp =145
\end{array}
\end{equation*}
Algunas Sumatorias Básicas:
\(\overset{n}{\underset{i=0}{\sum}}i=\frac{n(n+1)}{2}\text{.}\)
\(\overset{n}{\underset{i=0}{\sum}}i^{2}=\frac{n(n+1)(2n+1)}{6}\text{.}\)
\(\overset{n}{\underset{i=0}{\sum}}i^{3}=\left( \frac{n(n+1)}{2}\right)
^{2}\text{.}\)
\(\overset{n}{\underset{i=0}{\sum}}r^{i}=\frac{r^{n+1}-1}{r-1},
r\neq 0; r\neq 1 \text{.}\)
Proposición 2.3.3
Dada las siguientes sucesiones \(\{a_{n}\},\{b_{n}\}\) de números Reales y \(c\in\mathbb{R}\) entonces
\(\overset{n}{\underset{i=r}{\sum}}c\cdot a_i=c\overset{n}{\underset{i=r}
{\sum}}a_i\text{.}\)
\(\overset{s}{\underset{i=r}{\sum}}(a_{i}+b_{i})=\overset{s}{\underset
{i=r}{\sum}}a_{i}+\overset{s}{\underset{i=r}{\sum}}b_{i}\text{.}\)
\(\overset{s}{\underset{i=r}{\sum}}a_{i}=\overset{t}{\underset{i=r}{\sum}
}a_{i}+\overset{s}{\underset{i=t+1}{\sum}}a_{i}\) , donde \(r\leq \lt t
s\text{.}\)
\(\overset{s}{\underset{i=r}{\sum}}(a_{i}-a_{i+1})=a_{r}-a_{s+1}\text{,}\) propiedad telescópica.
) \(\overset{s}{\underset{i=r}{\sum}}c=(s-r+1)c\text{.}\)
\(\overset{s}{\underset{i=r}{\sum}}a_{i}=\overset{s-p}{\underset{i=r-p}
{\sum}}a_{i+p}\text{,}\) donde \(r-p\geq0\text{.}\)
Ejemplo 2.3.4
Calcular
\begin{equation*}
\overset{n}{\underset{i=0}{\sum}}(2i+1)\text{.}
\end{equation*}
Usando propiedades tenemos que
\begin{equation*}
\begin{array}{rl}
\overset{n}{\underset{i=0}{\sum}}(2i+1) \amp =\overset{n}{\underset{i=0}{\sum}
}2i+\overset{n}{\underset{i=0}{\sum}}1\\
\amp =2\overset{n}{\underset{i=0}{\sum}}i+\overset{n}{\underset{i=0}{\sum}}1\\
\amp =2\left( \frac{n(n+1)}{2}\right) +(n+1)\\
\amp =n^{2}+n+n+1\\
\amp =n^{2}+2n+1\\
\amp =(n+1)^{2}.
\end{array}
\end{equation*}
Ejemplo 2.3.5
Calcular
\begin{equation*}
\overset{n}{\underset{i=r}{\sum}}i
\end{equation*}
Usando propiedades tenemos que
\begin{equation*}
\begin{array}{rl}
\overset{n}{\underset{i=r}{\sum}}i \amp =\overset{n-r}{\underset{i=r-r}{\sum}
}(i+r)\\
\amp =\overset{n-r}{\underset{i=0}{\sum}}(i+r)\\
\amp =\overset{n-r}{\underset{i=0}{\sum}}i+\overset{n-r}{\underset{i=0}{\sum}
}r\\
\amp =\frac{(n-r)(n-r+1)}{2}+(n-r+1)r\\
\amp =\frac{(n-r+1)(n-r+2r)}{2}\\
\amp =\frac{(n-r+1)(n+r)}{2}.
\end{array}
\end{equation*}
Subsección Ejercicios
Calcular las siguientes sumatorias:
- \(\overset{100}{\underset{i=0}{\sum}}(i+7)^{2}\)
- \(\overset{n}{\underset{i=1}{\sum}}\frac{1}{r(r+1)}\)
- \(\overset{n}{\underset{k=1}{\sum}}\frac{1}{(k+1)!}\)
- \(\overset{n}{\underset{k=1}{\sum}}k3^{k}\)
Ayuda:
\begin{equation*}
(k+1)3^{k+1}-k3^{k}=2k3^{k}+3^{k+1}.
\end{equation*}
Ejemplo 2.3.6
Sea \(r\in\mathbb{R}-\mathbb{\{}0,1\mathbb{\}}\) Calcular \(\overset
{n}{\underset{k=0}{\sum}}kr^{k}\)
Veamos primero que, por propiedad telescópica se tiene que
\begin{equation*}
\overset{n}{\underset{k=0}{\sum}}\left[ \left( k+1\right) r^{k+1}
-kr^{k}\right] =\left( n+1\right) r^{n+1}
\end{equation*}
Reescribiendo las sumatoria tenemos
\begin{equation*}
\begin{array}{rl}
\overset{n}{\underset{k=0}{\sum}}\left[ \left( k+1\right) r^{k+1}
-kr^{k}\right] \amp =\sum_{k=0}^{n}\left[ \left( r-1\right) kr^{k}
+r^{k+1}\right] \\
\amp =\sum_{k=0}^{n}\left( r-1\right) kr^{k}+\sum_{k=0}^{n}r^{k+1}\\
\amp =\left( r-1\right) \sum_{k=0}^{n}kr^{k}+r\frac{r^{n+1}-1}{r-1}
\end{array}
\end{equation*}
Igualando los resultados tenemos
\begin{equation*}
\begin{array}{rl}
\overset{n}{\underset{k=0}{\sum}}\left[ \left( k+1\right) r^{k+1}
-kr^{k}\right] \amp =\overset{n}{\underset{k=0}{\sum}}\left[ \left(
k+1\right) r^{k+1}-kr^{k}\right] \\
\left( r-1\right) \sum_{k=0}^{n}kr^{k}+r\frac{r^{n+1}-1}{r-1} \amp =\left(
n+1\right) r^{n+1}\\
\sum_{k=0}^{n}kr^{k} \amp =\frac{\left( n+1\right) }{r-1}r^{n+1}
-r\frac{r^{n+1}-1}{\left( r-1\right)^{2}}
\end{array}
\end{equation*}
De lo cual se obtiene que
\begin{equation*}
\sum_{k=0}^{n}kr^{k} =\frac{ n r^{n+2} -(n+1)r^{n+1}+1}{left( r-1\right)^{2}}
\end{equation*}
Proposición 2.3.7
Dada las sucesiones \(\{a_{n}\}_{n\in\mathbb{N}}\) y \(\{b_{n}\}_{n\in\mathbb{N}
}\text{,}\) de Números Reales y \(c\in\mathbb{R}\)
- \(\overset{s}{\underset{k=r}{\prod}}(a_{k}\cdot b_{k})=\overset{s}
{\underset{k=r}{\prod}}a_{k}\cdot\overset{s}{\underset{k=r}{\prod}}b_{k}.\)
- \(\overset{s}{\underset{k=r}{\prod}}c\cdot a_{k}=c^{s-r+1}\cdot\overset
{s}{\underset{k=r}{\prod}}a_{k}.\)
- \(\overset{s}{\underset{k=r}{\prod}}c=c^{s-r+1}.\)
\(\overset{s}{\underset{k=r}{\prod}}a_{k}=\overset{t}{\underset{k=r}{\prod}
}a_{k}\cdot\overset{s}{\underset{k=t+1}{\prod}}a_{k}\text{,}\) donde \(r\leq t \lt s\text{.}\)
-
Propiedad Telescópica.
Si \((\forall k\in\mathbb{N})(a_{k}\not =0)\text{,}\) entonces
\begin{equation*}
\overset{s}{\underset{k=r}{\prod}}\frac{a_{k+1}}{a_{k}}=\frac{a_{s+1}}{a_{r}
}\text{.}
\end{equation*}
\(\overset{s}{\underset{k=r}{\prod}}a_{k}=\overset{s-p}{\underset
{k=r-p}{\prod}}a_{k+p}\text{.}\)
\(\overset{s}{\underset{k=r}{\prod}}c^{a_{k}}=c^{\overset{s}{\underset
{k=r}{\sum}}a_{k}}\text{,}\) con \(c\gt 0\)
Demostración
Por inducción demostraremos la propiedad telescópica. Se define
\begin{equation*}
p(n):\overset{n}{\underset{k=0}{\prod}}\frac{a_{k+1}}{{a_{k}}}{=}\frac{a_{n+1}
}{a_{0}}.
\end{equation*}
Verifiquemos \(p(0)\)
\begin{equation*}
\begin{array}{rl}
p(0):\overset{0}{\underset{k=0}{\prod}}\frac{a_{k+1}}{a_{k}} \amp =\frac{a_{1}
}{a_{0}}\\
\frac{a_{0+1}}{a_{0}} \amp =\frac{a_{1}}{a_{0}}\\
\frac{a_{1}}{a_{0}} \amp =\frac{a_{1}}{a_{0}}.
\end{array}
\end{equation*}
Luego \(p(0)\) es verdadero. Ahora veamos
\begin{equation*}
(\forall n\in\mathbb{N})[ p(n) \Rightarrow p(n+1)]
\end{equation*}
es decir, hipótesis
\begin{equation*}
p(n):\overset{n}{\underset{k=0}{\prod}}\frac{a_{k+1}}{a_{k}}=\frac{a_{n+1}
}{a_{0}},
\end{equation*}
Tesis:
\begin{equation*}
p(n+1):\overset{n+1}{\underset{k=0}{\prod}}\frac{a_{k+1}}{a_{k}}=\frac
{a_{n+2}}{a_{0}},
\end{equation*}
entonces
\begin{equation*}
\begin{array}{rl}
\overset{n+1}{\underset{k=0}{\prod}}\frac{a_{k+1}}{a_{k}} \amp = \overset{n}{\underset{k=0}{\prod}}\frac{a_{k+1}}{a_{k}} \cdot\left( \frac{a_{n+1+1}}{a_{n+1}}\right) \\
\amp =\left(
\frac{a_{n+1}}{a_{0}}\right) \cdot\left( \frac{a_{n+2}}{a_{n+1}}\right) \\
\amp =\frac{a_{n+1}}{a_{0}}\cdot\frac{a_{n+2}}{a_{n+1}}\\
\amp =\frac{a_{n+2}}{a_{0}}.
\end{array}
\end{equation*}
Así tenemos la segunda parte y por teorema de inducción concluimos
\begin{equation*}
(\forall n\in\mathbb{N})\left( \overset{n}{\underset{k=0}{\prod}}
\frac{a_{k+1}}{a_{k}}=\frac{a_{n+1}}{a_{0}}\right)
\end{equation*}
con \(a_{k}\not =0\text{,}\) \(\forall k\in\mathbb{N}\text{.}\)
Subsección Ejercicios
Demostrar por inducción la propiedad telescópica,
\begin{equation*}
(\forall n\in\mathbb{N})\left( \overset{n}{\underset{k=0}{\sum}}
(a_{k+1}-a_{k})=a_{n+1}-a_{0}\right) .
\end{equation*}
Ejemplo 2.3.8
Calcular
\begin{equation*}
\overset{s}{\underset{k=3}{\prod}}3^{(k^{2})}
\end{equation*}
Por la propiedad 5 se tiene que
\begin{equation*}
\begin{array}{rl}
\overset{7}{\underset{k=3}{\prod}}3^{(k^{2})} \amp =3^{3^{2}}\cdot3^{4^{2}
}\cdot\cdot\cdot3^{7^{2}}\\ \amp
=3^{(3^{2}+4^{2}+\cdot\cdot\cdot+7^{2})}\\ \amp
=3^{\overset{7}{\underset{k=3}{\sum}}k^{2}} =3^{140-5}=3^{135}.
\end{array}
\end{equation*}
