Question

Evaluate the infinite series. Round to the nearest tenth of needed

Answer 1

... and the series is?

Answer 2

formula

Answer 3

Do you see how
\[\sum_{n=1}^\infty 3\left(\frac{1}{4}\right)^{n-1}=\sum_{n=0}^\infty 3\left(\frac{1}{4}\right)^n~~?\]
The sum of the series would then be \(\dfrac{a}{1-r}\) provided that \(|r|<1\).

For this particular series, \(a=3\) and \(r=\dfrac{1}{4}\).

Answer 4

That's not what I'm getting.
\[\frac{3}{1-\dfrac{1}{4}}=\frac{3}{\dfrac{3}{4}}=3\cdot\frac{4}{3}=4\]

Answer 5

Yeah I see what I did wrong

Answer 6

thankyou

Answer 7

You're welcome