Question

Conjecture uma fórmula fechada para o somatório $\displaystyle\sum_{i=1}^{n} \dfrac{1}{i\cdot(i+1)}$ e prove-a.

Answer 1

Temos a seguinte soma:

$\displaystyle\sum_{i=1}^{n} \dfrac{1}{i\cdot(i+1)} = \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot 3}+\cdots +\dfrac{1}{(n-1)\cdot n}+\dfrac{1}{n\cdot(n+1)}$

Observe que:

$\dfrac{1}{x\cdot(x+1)} = \dfrac{1+0}{x\cdot(x+1)} = \dfrac{1+x-x}{x\cdot (x+1)} = \dfrac{x+1}{x\cdot(x+1)}-\dfrac{x}{x\cdot(x+1)} = \dfrac{1}{x}-\dfrac{1}{x+1}$

Aplicando isto na soma:

$\displaystyle\sum_{i=1}^{n} \dfrac{1}{i\cdot(i+1)} = \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot 3}+\cdots +\dfrac{1}{(n-1)\cdot n}+\dfrac{1}{n\cdot(n+1)}$

$\displaystyle\sum_{i=1}^{n} \left(\dfrac{1}{i} -\dfrac{1}{i+1}\right)=\left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots +\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)$

'Cortando' os termos semelhantes:

$\displaystyle\sum_{i=1}^{n} \left(\dfrac{1}{i} -\dfrac{1}{i+1}\right)=1-\dfrac{1}{n+1}$

$\displaystyle\sum_{i=1}^{n} \left(\dfrac{1}{i} -\dfrac{1}{i+1}\right)=\dfrac{n+1-1}{n+1}$

$\displaystyle\sum_{i=1}^{n} \left(\dfrac{1}{i} -\dfrac{1}{i+1}\right)=\dfrac{n}{n+1}$

$\boxed{\displaystyle\sum_{i=1}^{n} \dfrac{1}{i\cdot(i+1)}=\dfrac{n}{n+1}}$

Já está provada. Mas vou provar novamente com PIF.

Provando que para n =1 é válido:

$\displaystyle\sum_{i=1}^{1} \dfrac{1}{i\cdot(i+1)}=\dfrac{1}{2}$

De fato, essa soma equivale a $\dfrac{1}{2}$.

Assumimos que é válido para k. Provando que é válido para k+1.

$\displaystyle\sum_{i=1}^{k+1} \dfrac{1}{i\cdot(i+1)}=\dfrac{k+1}{k+2}$

$\displaystyle\sum_{i=1}^{k} \dfrac{1}{i\cdot(i+1)}+\dfrac{1}{(k+1)\cdot(k+2)}=\dfrac{k+1}{k+2}$

$\dfrac{k}{k+1}+\dfrac{1}{(k+1)\cdot(k+2)}=\dfrac{k+1}{k+2}$

$\dfrac{k}{k+1}+\left(\dfrac{1}{k+1}-\dfrac{1}{k+2}\right)=\dfrac{k+1}{k+2}$

$\dfrac{k+1}{k+1}=\dfrac{k+2}{k+2}$

$1= 1$

Como 1 =1, temos que essa propriedade é válida.