How to graph #\sum_{n=0}^\oo 2x^n#?

Question

Ezra Adams · Accepted Answer

Suppose that #sum_(n=0)^oo2x^n=S#. First, this series is only convergent when #-1< x<1#, as an infinite sum which summation terms does not tend to #0# must be divergent. So, #lim_(n->oo)x^n# exists if and only if #-1< x<1# Then,
#2+sum_(n=1)^oo2x^n=S#
#2+xsum_(n=1)^oo2x^(n-1)=S#
#2+xsum_(n=0)^oo2x^n=S# Since #S=sum_(n=0)^oo2x^n#,
#2+xS=S# Solving for #S#, we obtain #S=2/(1-x)#, which is only valid when #-1< x<1# (else the initial infinite sum is undefined). With this, we can easily graph the value of #sum_(n=0)^oo2x^n# for different values of #x#:
 graph{2/(1-x)sqrt(-(x+1)(x-1))/sqrt(-(x+1)(x-1)) [-2, 2, -1, 10]}

Charles Arocha · Answer

undefined To graph the function $ \sum_{n=0}^\infty 2x^n $, follow these steps:

1. Recognize that this function represents a geometric series.
2. The formula for the sum of an infinite geometric series is $ S = \frac{a}{1 - r} $, where $ a $ is the first term and $ r $ is the common ratio.
3. In this case, $ a = 2 $ and $ r = x $.
4. However, for $ |x| < 1 $, the series converges. Therefore, consider the domain as $ |x| < 1 $.
5. Substitute the values into the formula to get $ S = \frac{2}{1 - x} $.
6. Now, plot the graph of $ S = \frac{2}{1 - x} $ for $ |x| < 1 $.

That's how you graph $ \sum_{n=0}^\infty 2x^n $.