Don't know how to even begin

[dollyayesha2345]

Evaluate each infinite series that has a sum.
$$\sum ^{\infty}_{n=1}5\left(\frac{2}{3}\right)^{n-1}$$
I genuinely have no clue how to even approach this question.

 

[Zermelo]

Have you learned about geometric series?

 

[dollyayesha2345]

Have you learned about geometric series?

yes, but don't remember a single thing!

 

[Deleted member 4993]

yes, but don't remember a single thing!

A good way to remember is to get the textbook and study the worked-out example problems. Two important parameters of a geometric series are the first term (a) and the common ratio (r).

What are the values of those parameters ( a & r ) in the given problem?

 

[dollyayesha2345]

After Youtubing a lil bit this is my try: @Subhotosh Khan & @Zermelo
is this correct?

 

[blamocur]

You seem to be on a right track, but : $5\left(\frac{2}{3}\right)^{n-1} \neq \left(\frac{10}{3}\right)^{n-1}$

 

[dollyayesha2345]

You seem to be on a right track, but : $5\left(\frac{2}{3}\right)^{n-1} \neq \left(\frac{10}{3}\right)^{n-1}$

Hence the infinite series have 1 sum?

 

[blamocur]

Hence the infinite series have 1 sum?

No, you've made another erroneous transformation there. I'd suggest a better transformation:
$$\sum_{n=1}^\infty 5\left(\frac{2}{3}\right)^{n-1} =5\sum_{n=1}^\infty \left(\frac{2}{3}\right)^{n-1}$$Can you figure out the value of the sum?

 

[dollyayesha2345]

No, you've made another erroneous transformation there. I'd suggest a better transformation:
$$\sum_{n=1}^\infty 5\left(\frac{2}{3}\right)^{n-1} =5\sum_{n=1}^\infty \left(\frac{2}{3}\right)^{n-1}$$Can you figure out the value of the sum?

you want me to multiply 5 by the answer of $$S_n=\frac{1}{1-2/3}$$ basically? (I'll solve it later tho and show)

 

[blamocur]

Basically yes.

 

[Dr.Peterson]

You did this:


Do you understand that you can't pull the 5 inside the parentheses because the exponent is applied to the fraction before the multiplication? $\left(5\cdot\frac{2}{3}\right)^{n-1}$ would be equal to $(5)^{n-1}\left(\frac{2}{3}\right)^{n-1}$, not $5\left(\frac{2}{3}\right)^{n-1}$.

Moreover, even if you did have $\left(5\cdot\frac{2}{3}\right)^{n-1}$, what's iniside would be a multiplication $5\cdot\frac{2}{3}=\frac{10}{3}$, not a mixed number $5\frac{2}{3}=\frac{17}{3}$.

 

[dollyayesha2345]

Basically yes.

I believe this is correct

 

[blamocur]

I believe this is correctView attachment 29729

I agree that this infinite series converges to 5.

 

[pka]

I believe this is correctView attachment 29729

I agree that this infinite series converges to 5.

Actually $\displaystyle \sum\limits_{n = 1}^\infty {{{\left( {\frac{2}{3}} \right)}^{n - 1}}} = \frac{1}{{1 - \frac{2}{3}}} = 3$ so that $\displaystyle \sum\limits_{n = 1}^\infty {5{{\left( {\frac{2}{3}} \right)}^{n - 1}} = 15}$

 

[Steven G]

You just can't multiply the denominator by 3/1 as that will change the value of the fraction. You can, if you choose, multiply the numerator and denominator by 3.

You do know that arithmetic and algebra is a prerequisite for Calculus?

 

[blamocur]

Actually $\displaystyle \sum\limits_{n = 1}^\infty {{{\left( {\frac{2}{3}} \right)}^{n - 1}}} = \frac{1}{{1 - \frac{2}{3}}} = 3$ so that $\displaystyle \sum\limits_{n = 1}^\infty {5{{\left( {\frac{2}{3}} \right)}^{n - 1}} = 15}$

My mistake -- sorry