# Exercises on Geometric Series
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/geometric-series/
Fetched from algebrica.org test 16794; source modified 2025-05-12T22:58:39.
## Determine the nature of the following series and compute their sum.
-
\\[
\\text{1. } \\quad \\sum\_{n=1}^{\\infty} \\left( \\sqrt{3} \\right)^n
\\]
[solution](#s1)
-
\\[
\\text{2. } \\quad \\sum\_{n=1}^{\\infty} \\left( 3k + 2 \\right)^n
\\]
[solution](#s2)
-
\\[
\\text{3. } \\quad \\sum\_{n=1}^{\\infty} \\left( 1 - 2 \\cos x \\right)^n
\\]
[solution](#s3)
##### The proposed series are carefully designed to help you consolidate your understanding of [geometric series](../geometric-series.md). Try analyzing their behavior and computing their sums on your own before checking the provided solutions.
## Exercise 1
Determine the nature of the following series and compute its sum.
\\[
\\sum\_{n=1}^{\\infty} \\left( \\sqrt{3} \\right)^n
\\]
* * *
The given series is a geometric series with common ratio \\( \\sqrt{3} \\). A geometric series converges if and only if the common ratio satisfies:
\\[
|r| < 1
\\]
In this case we have
\\[
r = \\sqrt{3} \\approx 1.732 > 1
\\]
Therefore, in this case, the series diverges.
## Exercise 2
Determine the nature of the following series and compute its sum.
\\[
\\sum\_{n=1}^{\\infty} \\left( 3k + 2 \\right)^n
\\]
* * *
The proposed series is a geometric series whose common ratio \\( 3k + 2 \\) depends on the parameter \\( k \\). Let us recall that a geometric series converges whenever its common ratio \\( r \\) satisfies \\( |r| < 1 \\). Therefore, the condition for convergence is:
\\[
|3k + 2| < 1
\\]
We rewrite the [absolute value](../absolute-value.md) inequality as a double inequality:
\\[
-1 < 3k + 2 < 1
\\]
Solving for \\(k\\), we have:
\\[
-1 < 3k + 2 < 1
\\]
Subtract \\(2\\) from all parts and divide by \\(3\\). We obtain:
\\[
-3 < 3k < -1
\\]
\\[
-1 < k < -\\frac{1}{3}
\\]
Therefore, the series converges when \\(k\\) satisfies the above inequality. In this case, since the series starts at index \\(1\\), the formula to compute the sum is:
\\[
\\sum\_{n=1}^{\\infty} r^n = \\frac{r}{1 - r}, \\quad \\text{for } |r| < 1
\\]
In the given case, the sum is:
\\[
S = \\frac{3k + 2}{1 - (3k + 2)} = \\frac{3k + 2}{-3k - 1}
\\]
* * *
Let us now determine the values for which the series diverges. A geometric series diverges when \\( |r| \\geq 1 \\). In this case, we simply need to impose:
\\[
|3k + 2| \\geq 1
\\]
which can be rewritten as:
\\[
3k + 2 \\leq -1 \\quad \\text{or} \\quad 3k + 2 \\geq 1
\\]
From the previous inequalities, we conclude that the series is indeterminate for \\( k \\leq -1 \\), while it diverges for \\( k \\geq -\\dfrac{1}{3} \\).
To summarize:
- the series converges for:
\\[
-1 < k < -\\frac{1}{3} \\quad S = \\frac{3k + 2}{-3k - 1}
\\]
- The series diverges for \\( k \\geq -\\dfrac{1}{3} \\).
- The series is indeterminate for \\( k \\leq -1 \\).
## Exercise 3
Determine the nature of the following series and compute its sum.
\\[
\\sum\_{n=1}^{\\infty} \\left( 1 - 2 \\cos x \\right)^n
\\]
* * *
The proposed series is a geometric series whose common ratio \\(1 - 2\\cos x.\\) The series involves a [cosine function](../sine-and-cosine.md), so its convergence analysis must take into account the periodic nature of the trigonometric function. By applying the convergence criterion for geometric series, we impose ( |r| < 1 ):
\\[
|1 - 2\\cos x| < 1
\\]
This gives the inequality:
\\[
-1 < 1 - 2\\cos x < 1
\\]
Solving it, we obtain:
\\[
0 < \\cos x < 1
\\]
Therefore, the series converges for:
\\[
x \\in \\left(2k\\pi - \\frac{\\pi}{2}, 2k\\pi\\right), \\quad \\text{with } k \\in \\mathbb{Z}
\\]
Its sum is:
\\[
S = \\frac{1 - 2\\cos x}{1 - (1 - 2\\cos x)} = \\frac{1 - 2\\cos x}{2\\cos x}
\\]
## Glossary
- [Geometric series](../geometric-series.md): a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
- Common ratio \\( r \\): the constant factor by which each term in a geometric series is multiplied to get the next term.
- Convergence: the property of an infinite series where the sequence of its partial sums approaches a finite limit.
- Divergence: the property of an infinite series where the sequence of its partial sums does not approach a finite limit (it may go to infinity, negative infinity, or oscillate).
- Indeterminate: in the context of infinite series, a series that neither converges nor diverges to positive or negative infinity.
- Sum of a geometric series: the finite value that a convergent infinite geometric series approaches as the number of terms goes to infinity. For a series starting at index 1:
\\[
\\sum\_{n=1}^{\\infty} r^n = \\frac{r}{1 - r}, \\quad \\text{for } |r| < 1
\\]
- [Absolute value](../absolute-value.md): the non-negative value of a real number, ignoring its sign.
