Exercises on Geometric Series

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Determine the nature of the following series and compute their sum.

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\text{1. } \quad \sum_{n=1}^{\infty} \left( \sqrt{3} \right)^n

solution

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\text{2. } \quad \sum_{n=1}^{\infty} \left( 3k + 2 \right)^n

solution

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\text{3. } \quad \sum_{n=1}^{\infty} \left( 1 - 2 \cos x \right)^n

solution

The proposed series are carefully designed to help you consolidate your understanding of geometric series. 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 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, 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: 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: the non-negative value of a real number, ignoring its sign.