Resposta:

Primeiro calculamos a razão q:

\begin{gathered}q= \frac{a_n}{a_{n-1}}, n > 1 \\\\ q= \frac{a_2}{a_{2-1}} \\\\ q= \frac{a_2}{a_1} \\ \\ q= \frac{15}{45} \\ \\ q= \frac{1}{3}\end{gathered}

q=

a

n−1

a

n

,n>1

q=

a

2−1

a

2

q=

a

1

a

2

q=

45

15

q=

3

1

Agora, sabendo que a razão -1 > q > 1, podemos utilizar a fórmula da soma dos termos infinitos de uma P.G.:

\begin{gathered} \lim_{n \to \infty} S_n = \frac{a_1}{1-q} \\ \\ \lim_{n \to \infty} S_n= \frac{45}{1-( \frac{1}{3}) } \\ \\ \lim_{n \to \infty} S_n = \frac{45}{ \frac{3}{3}- \frac{1}{3} } \\ \\ \lim_{n \to \infty} S_n= \frac{45}{ \frac{2}{3} } \\ \\ \lim_{n \to \infty} S_n=45* \frac{3}{2} \\ \\ \lim_{n \to \infty} S_n= \frac{135}{2} \\ \\ \lim_{n \to \infty} S_n=67,5 \\ \\ \\ Renato.\end{gathered}

n→∞

lim

S

n

=

1−q

a

1

n→∞

lim

S

n

=

1−(

3

1

)

45

n→∞

lim

S

n

=

3

3

−

3

1

45

n→∞

lim

S

n

=

3

2

45

n→∞

lim

S

n

=45∗

2

3

n→∞

lim

S

n

=

2

135

n→∞

lim

S

n

=67,5

Renato.