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Use the properties of infinite series to evaluate the following series. $$\sum_{k=1}^{\infty}\left[\frac{1}{3}\left(\frac{5}{6}\right)^{k}+\frac{3}{5}\left(\frac{7}{9}\right)^{k}\right]$$

Short Answer

Expert verified
$$ Answer: The sum of the given infinite series is $$\frac{47}{10}$$.

Step by step solution

01

Formulate Sum of Two Geometric Series

Given that the infinite series can be split into two separate geometric series, let's write it as follows: $$S = \sum_{k=1}^{\infty}\frac{1}{3}\left(\frac{5}{6}\right)^{k}+\sum_{k=1}^{\infty}\frac{3}{5}\left(\frac{7}{9}\right)^{k}$$ We have now two different geometric series.
02

Identify the First Term and Common Ratio for Each Series

For the first series, we have the first term \(a_1=\frac{1}{3}\) and the common ratio \(r_1=\frac{5}{6}\). For the second series, we have the first term \(a_2=\frac{3}{5}\) and the common ratio \(r_2=\frac{7}{9}\).
03

Check Convergence of Each Series

Both series are geometric, hence they converge if the absolute value of their common ratios is less than 1. For the first series, \(|-r_1| = \left|\frac{5}{6}\right|<1\) For the second series, \(|-r_2| = \left|\frac{7}{9}\right|<1\) Both series converge, so we can proceed to find their sums.
04

Evaluate the First Series

Using the formula \(S = \frac{a}{1-r}\) for the first series, we get: $$S_1 = \frac{\frac{1}{3}}{1-\frac{5}{6}}=\frac{1/3}{1/6}=\frac{1}{3}\cdot\frac{6}{1}=2$$
05

Evaluate the Second Series

Using the formula \(S = \frac{a}{1-r}\) for the second series, we get: $$S_2 = \frac{\frac{3}{5}}{1-\frac{7}{9}}=\frac{3/5}{2/9}=\frac{3}{5}\cdot\frac{9}{2}=\frac{27}{10}$$
06

Add Both Series Results

Now we just need to add the results from both series to find the sum of the original series: $$S = S_1 + S_2 = 2 + \frac{27}{10} = \frac{20}{10}+\frac{27}{10}=\frac{47}{10}$$ So, the sum of the given infinite series is: $$\sum_{k=1}^{\infty}\left[\frac{1}{3}\left(\frac{5}{6}\right)^{k}+\frac{3}{5}\left(\frac{7}{9}\right)^{k}\right]=\frac{47}{10}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series
An infinite series is essentially a sum of infinitely many terms. When dealing with sequences and series, particularly those of infinite length, it's crucial to understand what these terms represent and how they are summed. An infinite series is generally represented in the form \( \sum_{k=1}^{\infty} a_k \), where \( a_k \) is the individual terms of the sequence.
A series can sometimes be extremely complex depending on the terms involved, but often in mathematics, we are interested in whether the sum approaches a specific value as more terms are added. This is where concepts like convergence come into play, helping us determine if the series can be assigned a finite sum value despite having infinitely many terms. For geometric series, which are a special type of infinite series, they converge under certain conditions. Let's explore this more in the next section.
Convergence
For an infinite series to be meaningful, mathematicians need to discuss its convergence. Convergence refers to the idea that as we keep adding terms from a series, the sum approaches a specific, fixed number. If the terms of a series do not approach a specific limit, the series is said to be divergent.
  • A series converges if the sequence of its partial sums tends to a limit, as the number of terms goes toward infinity.
  • For geometric series, convergence is straightforward to check: If the absolute value of the common ratio \(|r|\) is less than 1, then the series converges. Otherwise, it diverges.
  • In our original exercise, both series had common ratios of \(\frac{5}{6}\) and \(\frac{7}{9}\), which are less than 1, confirming that each series converges.
Understanding convergence is key when working with infinite series because it informs us whether we can meaningfully sum the series to get a finite result. Convergent series are not only meaningful but are also manageable with specific formulas to find their sums, which we'll see in the final section.
Sum of Geometric Series
When dealing with geometric series, there's a very handy formula to find their sum, as long as the series converges. For a geometric series with a first term \(a\) and a common ratio \(r\), the sum of the infinite series is given by:
\[S = \frac{a}{1 - r}\]This formula arises from the property that the series terms multiply over each other in a consistent manner, which makes it easier to handle them in a summation context.
  • The first term \(a\) is the term with which the sequence begins.
  • The formula is applicable, provided \(|r|<1\), ensuring convergence.
  • Using this formula, we calculated the sums of each component series separately in our original problem to reach a final result of \(\frac{47}{10}\).
This method of using the geometric series sum formula simplifies the computation significantly and is a powerful tool when addressing infinite sums of geometric sequences. It's important to confirm that the common ratio condition holds before applying the formula to ensure a valid result.

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Most popular questions from this chapter

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=4 a_{n}\left(1-a_{n}\right) ; a_{0}=0.5, n=0,1,2, \dots$$

Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than \(10^{-4} .\) Although you do not need it, the exact value of the series is given in each case. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2 k+1) !}$$

Determine whether the following series converge or diverge. $$\sum_{k=1}^{\infty} \frac{2^{k}+3^{k}}{4^{k}}$$

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

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