Respuesta :
Answer:
Check step-by-step-explanation.
Step-by-step explanation:
A given criteria for geometric series of the form [tex]\sum_{n=0}^{\infty} r^n[/tex] is that [tex]|r|<1[/tex]. Other wise, the series diverges. When it converges, we know that
[tex] \sum_{n=0}^\infty r^n = \frac{1}{1-r}[/tex].
So,
a)[tex]\sum_{n=0}^\infty (\frac{3}{2})^n[/tex] diverges since [tex]\frac{3}{2}>1[/tex]
b)[tex]\sum_{n=0}^\infty (\frac{1}{2})^n [/tex]converges since [tex]\frac{1}{2}<1[/tex], and
[tex]\sum_{n=0}^\infty (\frac{1}{2})^n= \frac{1}{1-\frac{1}{2}} = \frac{2}{2-1} = 2[/tex]
c)We can use the series in b) but starting at n=1 instead of n=0. Since they differ only on one term, we know it also converges and
[tex]\sum_{n=1}^{\infty}(\frac{1}{2})^n = \sum_{n=0}^{\infty}(\frac{1}{2})^n-(\frac{1}{2})^0 = 2-1 = 1[/tex].
d)Based on point c, we can easily generalize that if we consider the following difference
[tex]\sum_{n=1}^\infty r^n-\sum_{n=0}^\infty r^n = r^0 = 1[/tex]
So, they differ only by 1 if the series converges.
