Question

Como faço essa questão: Somatorio de n+1 ao + infinito de (-1) elevado a n+1 vezes 3/2^n.

Answer 1

$\Sigma^{n+1}_{+\infty} \ \frac{(-1)^{n+1}.3}{2^{n}}$


Usaremos regra do quociente ... 

Se n < 1 converge 
Se n > 1 diverge 
Se n = 1 indeterminado 

Teremos para o quociente : 

 an  -----> encostaremos o + 1 a cada n  
------
 bn ------> repete a equação 

Teremos então : 

$\Sigma^{n+1}_{+\infty} \ |\ \frac{\frac{(-1)^{n+1+1}.3}{2^{n+1}}}{\frac{(-1)^{n+1}.3}{2^{n}}}|\\\\\\\Sigma^{n+1}_{+\infty} \ |\frac{ \frac{(-1)^{n}.(-1)^{2}.3}{2^{n}.2}}{ \frac{(-1)^{n}.(-1).3}{2^{n}}}|\\\\\\\Sigma^{n+1}_{+\infty} \ | \frac{\frac{(-1)^{n}.3}{2^{n}.2}}{ \frac{(-1)^{n}.-3}{2^{n}}} |$

Inverto e multiplico ... 

$\Sigma^{n+1}_{+\infty} \ \frac{(-1)^{n}.3}{2^{n}.2}.\ \frac{2^{n}}{{(-1)^{n}.-3}}\\\\\\(corto\ os\ semelhantes)\\\\\\\Sigma^{n+1}_{+\infty} \ \frac{3}{2.(-3)}\\\\\\\Sigma^{n+1}_{+\infty} \ \frac{3}{-6}\\\\\\\Sigma^{n+1}_{+\infty} \ \ \ \boxed{\boxed{ -\ \frac{1}{2}}}\\\\\\Como\ - \frac{1}{2} \ \textless \ \ 1 \ \ \ converge\ para\ -\frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$