Answer:
C
Step-by-step explanation:
A geometric series will only converge if - 1 < r < 1
sum to infinity = [tex]\frac{a}{1-r}[/tex]
The nth term formula for a geometric series is
[tex]a_{n}[/tex] = a[tex]r^{n-1}[/tex]
where a is the first term and r the common ratio
The only summation with - 1 < r < 1 is C where r = - 0.2
Answer: The correct option is
(C) [tex]\sum_{n=1}^{\infty}4(-0.2)^{n-1}.[/tex]
Step-by-step explanation: We are give to select the geometric series that converges.
We know that
the general (n-th) term of a common geometric series is given by
[tex]a_n=ar^{n-1}.[/tex]
And the series converges if the modulus of the common ratio is less than 1, .e., |r| < 1.
Now, for the first infinite geometric series, we have
[tex]a_n=\dfrac{2}{3}(-3)^{n-1}.[/tex]
So, the common ratio will be
[tex]r=-3~~~\Rightarrow |r|=3>1.[/tex]
That is, the series will not converge. Option (A) is incorrect.
For the second geometric series, we have
[tex]a_n=5(-1)^{n-1}.[/tex]
So, the common ratio will be
[tex]r=-1~~~\Rightarrow |r|=1.[/tex]
That is, the series will not converge. Option (B) is incorrect.
For the third geometric series, we have
[tex]a_n=4(-0.2)^{n-1}.[/tex]
So, the common ratio will be
[tex]r=-0.2~~~\Rightarrow |r|=0.2<1.[/tex]
That is, the series will CONVERGE. Option (C) is correct.
For the fourth geometric series, we have
[tex]a_n=0.6(-2)^{n-1}.[/tex]
So, the common ratio will be
[tex]r=-2~~~\Rightarrow |r|=2>1.[/tex]
That is, the series will not converge. Option (D) is incorrect.
Thus, (C) is the correct option.
