Answer:
1st problem:
Converges to 6
2nd problem:
Converges to 504
Step-by-step explanation:
You are comparing to [tex]\sum_{k=1}^{\infty} a_1(r)^{k-1}[/tex]
You want the ratio r to be between -1 and 1.
Both of these problem are so that means they both have a sum and the series converges to that sum.
The formula for computing a geometric series in our form is [tex]\frac{a_1}{1-r}[/tex] where [tex]a_1[/tex] is the first term.
The first term of your first series is 3 so your answer will be given by:
[tex]\frac{a_1}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}=6[/tex]
The second series has r=1/6 and a_1=420 giving me:
[tex]\frac{420}{1-\frac{1}{6}}=\frac{420}{\frac{5}{6}}=420(\frac{6}{5})=504[/tex].
