Calculus Geometric series question

[sllyanders87]

Can you help me to solve step by step? I want to learn which steps i should follow.

 

[pka]

$\Large\sum\limits_{n = 1}^\infty {{{3\left( {\frac{{ - 1}}{4}} \right)}^n}}$
Note that $a=\dfrac{-3}{4}$ is the first term and $r=\dfrac{-1}{4}$ is the common ratio.

 

[Deleted member 4993]

View attachment 18696

Can you help me to solve step by step? I want to learn which steps i should follow.

Do you know the equation of the sum of 'n' terms for geometric series?

 

[HallsofIvy]

You could also factor out the common "3" to write this as $\displaystyle 3\sum_{n= 1}^\infty \left(\frac{1}{4}\right)^n= 3\left(\frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot\right)$.

Write $\displaystyle S= \frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot$ and factor out a 1/4: $\displaystyle S= \frac{1}{4}\left(1+ \frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot\right)= \frac{1}{4}\left(1+ S\right)$. So $\displaystyle S= \frac{1}{4}+ \frac{1}{4}S$, $\displaystyle \frac{3}{4}S= \frac{1}{4}$, $\displaystyle S= \frac{1}{3}$. And since the desired sum was 3S, the sum is $\displaystyle \frac{3}{4}$. And you shouldn't be thinking about "what steps to follow". You don't solve problems by following steps, you solve them by thinking about what the words and symbols in the problem mean.

 

[pka]

You could also factor out the common "3" to write this as $\displaystyle 3\sum_{n= 1}^\infty \left(\frac{1}{4}\right)^n= 3\left(\frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot\right)$.

Write $\displaystyle S= \frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot$ and factor out a 1/4: $\displaystyle S= \frac{1}{4}\left(1+ \frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot\right)= \frac{1}{4}\left(1+ S\right)$. So $\displaystyle S= \frac{1}{4}+ \frac{1}{4}S$, $\displaystyle \frac{3}{4}S= \frac{1}{4}$, $\displaystyle S= \frac{1}{3}$. And since the desired sum was 3S, the sum is $\displaystyle \frac{3}{4}$. And you shouldn't be thinking about "what steps to follow". You don't solve problems by following steps, you solve them by thinking about what the words and symbols in the problem mean.

Prof Ivey: It is an alternating series. Is it not? Maybe is missed something? but I see $\Large{\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{3}{{{4^n}}}}}$???

 

[sllyanders87]

Do you know the equation of the sum of 'n' terms for geometric series?

View attachment 18701

You could also factor out the common "3" to write this as $\displaystyle 3\sum_{n= 1}^\infty \left(\frac{1}{4}\right)^n= 3\left(\frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot\right)$.

Write $\displaystyle S= \frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot$ and factor out a 1/4: $\displaystyle S= \frac{1}{4}\left(1+ \frac{1}{4}+ \frac{1}{4^2}+ \frac{1}{4^3}+ \cdot\cdot\cdot\right)= \frac{1}{4}\left(1+ S\right)$. So $\displaystyle S= \frac{1}{4}+ \frac{1}{4}S$, $\displaystyle \frac{3}{4}S= \frac{1}{4}$, $\displaystyle S= \frac{1}{3}$. And since the desired sum was 3S, the sum is $\displaystyle \frac{3}{4}$. And you shouldn't be thinking about "what steps to follow". You don't solve problems by following steps, you solve them by thinking about what the words and symbols in the problem mean.

Prof Ivey: It is an alternating series. Is it not? Maybe is missed something? but I see $\Large{\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{3}{{{4^n}}}}}$???

Solved it thanks