Don't know how to even begin

[dollyayesha2345]

Evaluate each infinite series that has a sum.
[math]\sum ^{\infty}_{n=1}5\left(\frac{2}{3}\right)^{n-1}[/math]
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 : [imath]5\left(\frac{2}{3}\right)^{n-1} \neq \left(\frac{10}{3}\right)^{n-1}[/imath]

 

[dollyayesha2345]

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

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:
[math]\sum_{n=1}^\infty 5\left(\frac{2}{3}\right)^{n-1} =5\sum_{n=1}^\infty \left(\frac{2}{3}\right)^{n-1}[/math]Can you figure out the value of the sum?

 

[dollyayesha2345]

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

you want me to multiply 5 by the answer of [math]S_n=\frac{1}{1-2/3}[/math] 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? [imath]\left(5\cdot\frac{2}{3}\right)^{n-1}[/imath] would be equal to [imath](5)^{n-1}\left(\frac{2}{3}\right)^{n-1}[/imath], not [imath]5\left(\frac{2}{3}\right)^{n-1}[/imath].

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

 

[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 [imath]\displaystyle \sum\limits_{n = 1}^\infty {{{\left( {\frac{2}{3}} \right)}^{n - 1}}} = \frac{1}{{1 - \frac{2}{3}}} = 3[/imath] so that [imath]\displaystyle \sum\limits_{n = 1}^\infty {5{{\left( {\frac{2}{3}} \right)}^{n - 1}} = 15}[/imath]

 

[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 [imath]\displaystyle \sum\limits_{n = 1}^\infty {{{\left( {\frac{2}{3}} \right)}^{n - 1}}} = \frac{1}{{1 - \frac{2}{3}}} = 3[/imath] so that [imath]\displaystyle \sum\limits_{n = 1}^\infty {5{{\left( {\frac{2}{3}} \right)}^{n - 1}} = 15}[/imath]

My mistake -- sorry