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:

\(\displaystyle S_{\infty}=\lim_{n\to\infty}S_n\)

 

[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:

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

It is my understanding your sum is:

\(\displaystyle \sum_{k=1}^{\infty}a_n\)

In this case, you would need to write:

\(\displaystyle S_{\infty}=\frac{a}{1-r}-a=\frac{ar}{1-r}\)

 

[HOI]

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

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

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