Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

Question

3 years ago

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Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

, if possible.
a) This is a convergent geometric series. The sum is -5
b) This is a divergent geometric series. The sum is -5
c) This is a convergent geometric seties. The sum cannot be found.
b) This is a diveregent geometric series. The sum cannot be found.

Mathematics

2 answers:

const2013 [10] · Answer 1 · 2021-11-16T16:07:01+03:00

Answer: The correct option is

(d) This is a divergent geometric series. The sum cannot be found.

Step-by-step explanation: The given infinite geometric series is

$S=\sum_{i=1}^{\infty}15(4)^{i-1}.$

We are to identify whether the given geometric series is convergent or divergent. If convergent, we are to find the sum of the series.

We have the D' Alembert's ratio test, states as follows:

Let, $\sum_{i=1}^{\infty}a_i$ is an infinite series, with complex coefficients $a_i$ and we consider the following limit:

$L=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}.$

Then, the series will be convergent if L < 1 and divergent if L > 1.

For the given series, we have

$a_i=15(4)^{i-1},\\\\a_{i+1}=15(4)^i.$

So, the limit is given by

$L\\\\\\=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i-1}}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i}4^{-1}}\\\\\\=\dfrac{1}{4^{-1}}\\\\=4>1.$