Question

Consider the series.

∞∑n=1 (3^n+1 / 3n+1)

Does the series converge or diverge?

Select answers from the drop-down menus to correctly complete the statements.

Answer 1

Put 1

`\\ \rm\Rrightarrow \left((3^(1+1))/(3(1)+1)\right)`

`\\ \rm\Rrightarrow 3^2/4=9/4=2.2`

Put 2

`\\ \rm\Rrightarrow \left((3^(2+1))/(3(2)+1)\right)=3^3/7=27/7=3.8`

R is

• 3.8/2.2=19/11

Its greater than 1 .

• So diverge series.

Answer 2

`r=(12)/(7)`

series diverges

Explanation:

To find the common ratio (r) of a geometric series, divide the (n+1)th term by the nth term.

When n = 1:

`a_1=(3^(1+1))/(3(1)+1)=(3^2)/(4)=\frac94`

When n =2:

`a_2=(3^(2+1))/(3(2)+1)=(3^3)/(7)=(27)/(7)`

Therefore,

`r=(a_2)/(a_1)=((27)/(7))/((9)/(4))=(12)/(7)`

A series that converges has a finite limit. If |r| < 1, then the series will converge.

A series that diverges means either the partial sums have no limit or approach infinity. If |r| > 1 then the series diverges.

Therefore, as the limit of the series approaches infinity and it's r value is greater than 1, the series diverges.