Convergence Of Series And Trigonometric Identity Of Sin(3x) A Detailed Explanation

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Hey guys! Let's dive into the fascinating world of infinite series and explore the convergence of the given series: $\sum_{n=0}^{\infty} \frac{2^n+5}{3^n}$. This is a classic example where we can apply our knowledge of geometric series to determine whether it converges or diverges, and if it converges, what its sum is. To truly grasp this, we'll break it down step by step, making sure we understand each component.

First off, when we talk about absolute convergence, we're essentially asking if the sum of the absolute values of the terms in the series converges. In simpler terms, we want to know if the series still converges even if we make all the terms positive. For this particular series, all the terms are already positive, so we can directly analyze the convergence of the series as it is.

Now, let's rewrite the series to make it easier to work with. We can split the fraction into two parts:

$$\sum_{n=0}^{\infty} \frac{2^n+5}{3^n} = \sum_{n=0}^{\infty} \left(\frac{2^n}{3^n} + \frac{5}{3^n}\right)$$

Using the properties of summation, we can further split this into two separate series:

$$\sum_{n=0}^{\infty} \frac{2^n}{3^n} + \sum_{n=0}^{\infty} \frac{5}{3^n}$$

This gives us:

$$\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n + 5\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n$$

Here's where the geometric series concept comes into play. A geometric series has the form $\sum_{n=0}^{\infty} ar^n$, where a is the first term and r is the common ratio. A geometric series converges if the absolute value of r is less than 1 (|r| < 1), and its sum is given by $\frac{a}{1-r}$. If |r| ≥ 1, the series diverges.

Looking at our first series, $\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n$, we see that a = 1 (since $(\frac{2}{3})^0 = 1$) and r = $\frac{2}{3}$. Since |$\frac{2}{3}$| < 1, this geometric series converges. Its sum is:

$$\frac{1}{1 - \frac{2}{3}} = \frac{1}{\frac{1}{3}} = 3$$

For the second series, $5\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n$, we have a = 1 and r = $\frac{1}{3}$. Again, |$\frac{1}{3}$| < 1, so this geometric series also converges. Its sum is:

$$5 \cdot \frac{1}{1 - \frac{1}{3}} = 5 \cdot \frac{1}{\frac{2}{3}} = 5 \cdot \frac{3}{2} = \frac{15}{2}$$

Now, we add the sums of the two convergent series to find the sum of the original series:

$$3 + \frac{15}{2} = \frac{6}{2} + \frac{15}{2} = \frac{21}{2}$$

So, the series $\sum_{n=0}^{\infty} \frac{2^n+5}{3^n}$ converges absolutely, and its sum is $\frac{21}{2}$. The initial calculation provided, $\sum_{n=0}^{\infty} \frac{2^n+5}{3^n}=\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n}=\frac{6}{3}=2 \textless 1$, is a bit misleading. While it's true that the limit of the terms $\frac{2^n+5}{3^n}$ as n approaches infinity is 0 (which is a necessary condition for convergence), it doesn't tell us the sum of the series. We need to use the geometric series formula to find the actual sum, which we've now determined to be $\frac{21}{2}$.

In summary, by splitting the original series into two geometric series and applying the geometric series formula, we've clearly shown that the series converges absolutely and calculated its sum. Understanding geometric series is super important for dealing with these kinds of problems!

$\sum_{n=0}^{\infty} \frac{2^n+5}{3^n}=\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n}=\frac{6}{3}=2 \textless 1$

Okay, let’s break down this part of the original statement and clarify why it's a bit off. The initial line, $\sum_{n=0}^{\infty} \frac{2^n+5}{3^n}=\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n}$, seems to be equating the sum of an infinite series with the limit of its terms as n approaches infinity. This is a huge misconception in calculus and series analysis, and it’s crucial to understand why this is incorrect. Guys, pay close attention here!

The limit $\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n}$ helps us determine if the series might converge, but it doesn't tell us what the series converges to. To illustrate this, let’s dive a little deeper.

The correct interpretation of the limit is as follows: If $\lim_{n \rightarrow \infty} a_n \neq 0$, then the series $\sum_{n=0}^{\infty} a_n$ diverges. This is known as the Divergence Test. However, if $\lim_{n \rightarrow \infty} a_n = 0$, it doesn't automatically mean the series converges. It just means the test is inconclusive, and we need to use other tests to determine convergence.

Let's analyze the limit in question: $\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n}$. As n gets larger and larger, $2^n$ and $3^n$ both grow exponentially, but $3^n$ grows faster than $2^n$. The constant 5 becomes insignificant compared to the exponential terms as n approaches infinity. Therefore, we can rewrite the limit as:

$$\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n} = \lim _{n \rightarrow \infty} \frac{2^n}{3^n} = \lim _{n \rightarrow \infty} \left(\frac{2}{3}\right)^n$$

Since $\frac{2}{3}$ is a fraction less than 1, raising it to higher and higher powers will make it approach 0. Thus:

$$\lim _{n \rightarrow \infty} \left(\frac{2}{3}\right)^n = 0$$

So, the limit of the terms is indeed 0, which means the Divergence Test is inconclusive. The series might converge, but we need further analysis to confirm. The original statement incorrectly jumps to the conclusion that because the limit is less than 1 (it’s actually 0), the sum of the series is 2. The calculation $\frac{6}{3} = 2$ seems to come out of nowhere and has no valid mathematical basis in this context.

To reiterate, finding the limit of the terms only tells us if the terms are getting infinitesimally small, which is a prerequisite for convergence, but it doesn’t tell us the sum. Think of it like this: imagine adding smaller and smaller pieces of a pie. The pieces are getting tiny (approaching zero), but the total amount of pie you have might still add up to a finite amount, an infinite amount, or somewhere in between. That’s what the sum of the series represents.

To find the sum, we need to correctly apply techniques like recognizing geometric series, as we did earlier, and using the appropriate formulas. Mistaking the limit of the terms for the sum of the series is a common but critical error in understanding infinite series. So, always remember: limit to zero is a good sign, but it’s not the whole story!

In conclusion, the statement $\sum_{n=0}^{\infty} \frac{2^n+5}{3^n}=\lim _{n \rightarrow \infty} \frac{2^n+5}{3^n}=\frac{6}{3}=2 \textless 1$ is a flawed argument. We've established that the series does converge, but to $\frac{21}{2}$, not 2, and we arrived at this conclusion by correctly using the properties of geometric series, not by misinterpreting the limit of the terms.

$\sin(3x)=$

Let's switch gears now and dive into the world of trigonometry! We're going to explore the expansion of $\sin(3x)$. This is a fantastic example of how we can use trigonometric identities to express more complex trigonometric functions in terms of simpler ones. This skill is super useful in calculus, physics, and engineering, where simplifying expressions often makes problems much easier to solve. So, let's unravel this trigonometric mystery together!

The key here is to use the sum of angles formula for sine. This formula states that:

$$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$$

To find $\sin(3x)$, we can think of $3x$ as $2x + x$. So, we can apply the sum of angles formula like this:

$$\sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x)$$

Now, we need to deal with $\sin(2x)$ and $\cos(2x)$. For this, we'll use the double angle formulas. These formulas are derived from the sum of angles formulas themselves and are essential tools in trigonometry. The double angle formulas we need are:

$$\sin(2x) = 2\sin(x)\cos(x)$$

$$\cos(2x) = \cos^2(x) - \sin^2(x)$$

We could also use $\cos(2x) = 1 - 2\sin^2(x)$ or $\cos(2x) = 2\cos^2(x) - 1$, but the first form is usually the most straightforward to start with.

Let's substitute these double angle formulas back into our expression for $\sin(3x)$:

$$\sin(3x) = (2\sin(x)\cos(x))\cos(x) + (\cos^2(x) - \sin^2(x))\sin(x)$$

Now, we'll expand and simplify this expression:

$$\sin(3x) = 2\sin(x)\cos^2(x) + \sin(x)\cos^2(x) - \sin^3(x)$$

Combining the like terms (the terms with $\sin(x)\cos^2(x)$), we get:

$$\sin(3x) = 3\sin(x)\cos^2(x) - \sin^3(x)$$

Okay, we're getting closer, but we can simplify this even further! We want to express $\sin(3x)$ solely in terms of $\sin(x)$. To do this, we'll use the Pythagorean identity, which is one of the most fundamental identities in trigonometry:

$$\sin^2(x) + \cos^2(x) = 1$$

From this identity, we can express $\cos^2(x)$ as:

$$\cos^2(x) = 1 - \sin^2(x)$$

Now, substitute this into our expression for $\sin(3x)$:

$$\sin(3x) = 3\sin(x)(1 - \sin^2(x)) - \sin^3(x)$$

Expand and simplify again:

$$\sin(3x) = 3\sin(x) - 3\sin^3(x) - \sin^3(x)$$

Finally, combine the $\sin^3(x)$ terms:

$$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$

And there we have it! We've successfully expressed $\sin(3x)$ in terms of $\sin(x)$. This final form, $\sin(3x) = 3\sin(x) - 4\sin^3(x)$, is a crucial identity in trigonometry and is used in various mathematical and scientific applications.

So, to recap, we used the sum of angles formula and the double angle formulas to initially expand $\sin(3x)$. Then, we used the Pythagorean identity to express the entire equation in terms of $\sin(x)$. This methodical approach is key to tackling many trigonometric problems. By breaking down complex expressions into simpler components and applying the appropriate identities, we can often find elegant and useful solutions.

This entire discussion falls squarely into the realm of mathematics. Specifically, it touches on several key areas within mathematics:

• Calculus: The discussion of infinite series and their convergence is a fundamental topic in calculus. Understanding convergence and divergence is crucial for working with integrals, power series, and many other advanced concepts.
• Real Analysis: A deeper dive into the theory of infinite series and sequences falls under the domain of real analysis. This field rigorously studies the foundations of calculus and includes topics like limits, continuity, and convergence.
• Trigonometry: The exploration of $\sin(3x)$ and the use of trigonometric identities are core concepts in trigonometry. These identities are essential for simplifying trigonometric expressions, solving equations, and applications in various fields.
• Mathematical Analysis: This is a broad area encompassing calculus and real analysis, focusing on the rigorous treatment of mathematical concepts and problem-solving techniques.

So, whether you're a student delving into calculus for the first time or a seasoned mathematician exploring advanced concepts, this discussion offers valuable insights into the interconnectedness of different areas within mathematics. Keep exploring, keep questioning, and keep enjoying the beauty of mathematics!