MHB Geometric Series: Find Sum of Infinity - 9-32-n

[ChelseaL]

Given that the sum of the first n terms of series, s, is 9-3^2-n

Find the sum of infinity of s.

Do I use the formula S_\infty = \frac{a}{1-r}?

 

[tkhunny]

Given that the sum of the first n terms of series, s, is 9-3^2-n

Find the sum of infinity of s.

Do I use the formula S_\infty = \frac{a}{1-r}?

I can't tell what your geometric terms are. Are they supposed to be $9 - 3^{2-n}\;for\;n\in \mathbb{N}$? Are you SURE that's a geometric sequence?

 

[ChelseaL]

I'm not exactly sure what the topic is, but this is the part before it.
https://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/geometric-series-24025.html

 

[MarkFL]

What you can do is:

$$S_{\infty}=\lim_{n\to\infty}S_n$$

 

[ChelseaL]

How do I work it out from there?

 

[MarkFL]

How do I work it out from there?

Have you worked with limits before?

 

[ChelseaL]

This is my first time hearing this term. I don't think my lecturer expects us to use that method since we never covered it in class.

 

[MarkFL]

This is my first time hearing this term. I don't think my lecturer expects us to use that method since we never covered it in class.

Okay, then the formula you first cited works in this case:

$$S_{\infty}=\sum_{k=0}^{\infty}a_n=\frac{a}{1-r}$$

It is my understanding your sum is:

$$\sum_{k=1}^{\infty}a_n$$

In this case, you would need to write:

$$S_{\infty}=\frac{a}{1-r}-a=\frac{ar}{1-r}$$

 

[HOI]

Consider the finite geometric series [math]S= a+ at+ at^2+ at^3+ \cdot\cdot\cdot+ at^n[/math]. Subtract a from both sides: [math]S- a=at+ at^2+ at^3+ \cdot\cdot\cdot+ at^n[/math]. Factor "t" out of the right side: [math]S- a= t(a+ at+ at^2+ \cdot\cdot\cdot+ at^{n- 1})[/math]. The sum on the right is almost the same as the original sum- it is only missing the "[math]at^n[/math]" term. Add and subtract [math]at^n[/math] inside the parentheses:
[math]S- a= t(a+ at+ at^2+ \cdot\cdot\cdot+ at^{n- 1}+ at^n- at^n)= t(a+ at+ at^2+ \cdot\cdot\cdot+ at^{n- 1}+ at^n)- at^{n+1}= tS- at^{n+1}[/math]

Add [math]at^{n+1}[/math] to both sides:
[math]S- a+ at^{n+ 1}= tS[/math]
[math]S- tS= (1- t)S= a- at^{n+1}= a(1- t^{n+1})[/math]
Finally, divide both sides by 1- t:
[math]S= \frac{a(1- t^{n+1})}{1- t}[/math]

The sum of the infinite series, [math]\sum_{n=0}^\infty at^n[/math] is the limit of that as n goes to infinity. If [math]t\ge 1[/math], that limit, so that sum, does not exist. If [math]t<1[/math], [math]t^{n+ 1}[/math] goes to 0 so
[math]\sum_{n=0}^\infty at^n= \frac{a}{1- t}[/math].