Question

Evaluate the infinite series

Answer 1

@satellite73

Answer 2

pull the \(-3\) out front, it is a constant
then you have
\[-3\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^{n-1}\]

Answer 3

since \(\frac{3}{4}<1\) you sum the geometric series via
\[\sum r^n=\frac{1}{1-r}\]

Answer 4

you can just about to it in your head
one minus three fourths is one fourth
the reciprocal of one fourth is 4
then \(-3\times 4\) to finish

Answer 5

-12

Answer 6

?

Answer 7

yup