Question

Which geometric series converges ???

Answer 1

Answer:

C

Step-by-step explanation:

A geometric series will only converge if - 1 < r < 1

sum to infinity = $\frac{a}{1-r}$

The nth term formula for a geometric series is

$a_{n}$ = a$r^{n-1}$

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 2

Answer:  The correct option is

(C) $\sum_{n=1}^{\infty}4(-0.2)^{n-1}.$

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

$a_n=ar^{n-1}.$

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

$a_n=\dfrac{2}{3}(-3)^{n-1}.$

So, the common ratio will be

$r=-3~~~\Rightarrow |r|=3>1.$

That is, the series will not converge. Option (A) is incorrect.

For the second geometric series, we have

$a_n=5(-1)^{n-1}.$

So, the common ratio will be

$r=-1~~~\Rightarrow |r|=1.$

That is, the series will not converge. Option (B) is incorrect.

For the third geometric series, we have

$a_n=4(-0.2)^{n-1}.$

So, the common ratio will be

$r=-0.2~~~\Rightarrow |r|=0.2$

That is, the series will CONVERGE. Option (C) is correct.

For the fourth geometric series, we have

$a_n=0.6(-2)^{n-1}.$

So, the common ratio will be

$r=-2~~~\Rightarrow |r|=2>1.$

That is, the series will not converge. Option (D) is incorrect.

Thus, (C) is the correct option.