Practice Exercises

To find the limit of the sequence, we can divide both the numerator and denominator by the highest power of \(n\), which is \(n^2\):

Therefore, the sequence converges to \(\frac{3}{5}\).

Let's examine the behavior of this sequence:

The absolute value of the sequence approaches 1 as \(n\) approaches infinity:

However, the sequence alternates between positive and negative values due to the \((-1)^n\) term. Since the absolute value approaches 1 and the sign keeps changing, the sequence does not converge to a single value.

Therefore, the sequence diverges.

Let's examine the pattern in this sequence:

• \(a_1 = 2\)
• \(a_2 = 6\)
• \(a_3 = 18\)
• \(a_4 = 54\)
• \(a_5 = 162\)

Looking at the ratio between consecutive terms:

• \(\frac{a_2}{a_1} = \frac{6}{2} = 3\)
• \(\frac{a_3}{a_2} = \frac{18}{6} = 3\)
• \(\frac{a_4}{a_3} = \frac{54}{18} = 3\)
• \(\frac{a_5}{a_4} = \frac{162}{54} = 3\)

This is a geometric sequence with first term \(a_1 = 2\) and common ratio \(r = 3\).

The formula for the general term of a geometric sequence is \(a_n = a_1 \cdot r^{n-1}\).

Therefore, \(a_n = 2 \cdot 3^{n-1}\).

Let's rewrite the series:

This is similar to the form of a power series. Let's use the ratio test to determine convergence:

Since \(\left|\frac{a_{n+1}}{a_n}\right| = \frac{2}{3} < 1\), the series converges by the ratio test.

To find the sum, we can use the fact that this is related to the Taylor series for \(-\ln(1-x)\):

With \(x = \frac{2}{3}\), we get:

Therefore, the series converges to \(\ln(3)\).

Let's examine the behavior of the general term as \(n\) approaches infinity:

Since the limit of the general term is 0, the series might converge (this is a necessary but not sufficient condition for convergence).

Let's compare this series with a known series. For large values of \(n\), we have:

We know that \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) is a convergent p-series (with \(p = 2 > 1\)).

More formally, we can use the limit comparison test with \(b_n = \frac{1}{n^2}\):

Since this limit is finite and positive, and \(\sum b_n\) converges, by the limit comparison test, \(\sum a_n\) also converges.

Therefore, the series \(\sum_{n=1}^{\infty} \frac{n^2}{n^4 + 1}\) converges.

Let's rewrite the series:

This is a geometric series with first term \(a = 1\) and common ratio \(r = \frac{2}{5}\).

Since \(|r| = \frac{2}{5} < 1\), the series converges, and the sum is given by:

Therefore:

The sum of the series is 1.

Let's use the ratio test for this series. Let \(a_n = \frac{n!}{n^n}\).

Since \(\left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{e} < 1\), the series converges by the ratio test.

Let's use the integral test for this series. First we need to check the conditions of the integral test: the function \(f(x)=\frac{1}{x ln(x)}\) is positive, decreasing, and continous on the interval \([2, \infty)\). We can check that the function is positive for \(x \geq 2\) since both \(x\) and \(\ln(x)\) are positive in this interval.

Next, we need to check if the function is continuous. The function \(f(x)\) is defined and hence continuous for \(x > 1\) since \(\ln(x)\) is defined and positive in this interval.

Finally, we need to check if the function is decreasing. We can do this by taking the derivative:

Since \(\ln(x) + 1 > 0\) for \(x \geq 2\), we have \(f'(x) < 0\), which means that \(f(x)\) is decreasing.

Now we can apply the integral test. We need to evaluate the improper integral:

We need to check if the improper integral \(\int_{2}^{\infty} \frac{1}{x \ln(x)} \, dx\) converges.

Let \(u = \ln(x)\), then \(du = \frac{1}{x} \, dx\).

Now, let's evaluate the improper integral:

Since the improper integral diverges, by the integral test, the series \(\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}\) also diverges.

This is an alternating series of the form \(\sum_{n=1}^{\infty} (-1)^n b_n\) where \(b_n = \frac{1}{\sqrt{n}}\).

To apply the alternating series test, we need to check two conditions:

1. \(b_n > 0\) for all \(n\)
2. \(b_n\) is decreasing
3. \(\lim_{n \to \infty} b_n = 0\)

Checking these conditions:

1. \(b_n = \frac{1}{\sqrt{n}} > 0\) for all \(n \geq 1\). ✓
2. To check if \(b_n\) is decreasing, we need to verify that \(b_{n+1} \leq b_n\) for all \(n\):
   \begin{align} b_{n+1} &\leq b_n \\ \frac{1}{\sqrt{n+1}} &\leq \frac{1}{\sqrt{n}} \\ \sqrt{n} &\leq \sqrt{n+1} \end{align}
   This is true for all \(n \geq 1\), so \(b_n\) is decreasing. ✓
3. \(\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0\)
   ✓

Since all three conditions are satisfied, by the alternating series test, the series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\) converges.

Let's use the ratio test to find the radius of convergence. Let \(a_n = \frac{(x-3)^n}{n \cdot 2^n}\) for \(n \geq 1\) (we'll handle the \(n = 0\) term separately).

For convergence, we need \(\frac{|x-3|}{2} < 1\), which gives us:

So the radius of convergence is \(R = 2\).

Now we need to check the endpoints \(x = 1\) and \(x = 5\):

At \(x = 1\), the series becomes \(\sum_{n=0}^{\infty} \frac{(1-3)^n}{n \cdot 2^n} = \sum_{n=0}^{\infty} \frac{(-2)^n}{n \cdot 2^n} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n}\).

For \(n \geq 1\), this is the alternating harmonic series, which converges. The \(n = 0\) term is defined as 1, so it doesn't affect convergence.

At \(x = 5\), the series becomes \(\sum_{n=0}^{\infty} \frac{(5-3)^n}{n \cdot 2^n} = \sum_{n=0}^{\infty} \frac{2^n}{n \cdot 2^n} = \sum_{n=0}^{\infty} \frac{1}{n}\).

For \(n \geq 1\), this is the harmonic series, which diverges.

Therefore, the interval of convergence is \([1, 5)\).

We know that the power series for \(\frac{1}{1-x}\) is:

Taking the derivative of both sides:

Multiplying both sides by \(x\):

Therefore, \(f(x) = \frac{x}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n = x + 2x^2 + 3x^3 + 4x^4 + \ldots\) for \(|x| < 1\).

Let's rewrite the general term:

So we have:

We know from the previous exercise that:

To find \(\sum_{n=1}^{\infty} n^2 x^n\), we can take the derivative of \(\sum_{n=1}^{\infty} n x^n\) with respect to \(x\):

Multiplying both sides by \(x\):

Now we can find the sum of the original series:

Therefore, \(\sum_{n=1}^{\infty} n(n+1)x^n = \frac{2x}{(1-x)^3}\) for \(|x| < 1\).

To find the Taylor series centered at \(a = 1\), we need to compute the derivatives of \(f(x) = \ln(1+x)\) and evaluate them at \(x = 1\).

Using the Taylor series formula:

Therefore, the Taylor series for \(\ln(1+x)\) centered at \(a = 1\) is:

This series converges for \(|x-1| < 1\), or \(0 < x < 2\).

We'll use the Taylor series for \(\sin(x)\) centered at \(a = 0\):

By Taylor's Remainder Theorem, if we use the \(n\)th-degree Taylor polynomial \(P_n(x)\), the error is:

where \(c\) is some point between 0 and \(x\).

For \(\sin(x)\), the derivatives cycle through \(\sin(x)\), \(\cos(x)\), \(-\sin(x)\), and \(-\cos(x)\). Since these functions are bounded by 1 in absolute value, we have:

We need to find the smallest \(n\) such that:

Let's try some values:

• For \(n = 3\): \(\frac{|0.2|^4}{4!} = \frac{0.0016}{24} \approx 6.67 \times 10^{-5}\) (not small enough)
• For \(n = 5\): \(\frac{|0.2|^6}{6!} = \frac{0.000064}{720} \approx 8.89 \times 10^{-8}\) (small enough)

So we need to use the 5th-degree Taylor polynomial:

Therefore, \(\sin(0.2) \approx 0.19867\) with an error less than \(10^{-6}\).

The actual value of \(\sin(0.2) \approx 0.19867\), confirming our approximation.

We know the Maclaurin series for \(\frac{1}{1-x}\):

We can rewrite \(\frac{1}{1-x^2}\) as:

Substituting \(x^2\) for \(x\) in the known series:

Therefore, the Maclaurin series for \(f(x) = \frac{1}{1-x^2}\) is:

We'll use the Maclaurin series for \(e^x\) and substitute \(-x^2\) for \(x\):

Integrating term by term:

Evaluating the definite integral:

Let's compute the first few terms:

So we have:

To estimate the error, we need to bound the next term:

Since this is less than \(10^{-4}\), our approximation is accurate enough.

Therefore, \(\int_0^{0.5} e^{-x^2} \, dx \approx 0.4613\) with an error less than \(10^{-4}\).

Let's assume a solution of the form:

Taking the derivative:

Substituting into the differential equation \(y' + y = x\):

Comparing coefficients on both sides:

Using the initial condition \(y(0) = 1\), we have \(a_0 = 1\).

Therefore:

The solution is:

This is the power series solution to the differential equation \(y' + y = x\) with initial condition \(y(0) = 1\).

Note: The exact solution to this differential equation is \(y(x) = x - 1 + 2e^{-x}\), which can be verified by substituting back into the original equation.

This is a famous problem known as the Basel problem, first solved by Euler. We'll use the Fourier series approach.

Consider the function \(f(x) = x^2\) for \(-\pi \leq x \leq \pi\). The Fourier series for this function is:

where:

Using integration by parts twice:

Since \(f(x) = x^2\) is an even function, all \(b_n = 0\).

Therefore, the Fourier series is:

Evaluating at \(x = 0\):

We also know that:

Solving for \(\sum_{n=1, n \text{ odd}}^{\infty} \frac{1}{n^2}\):

Substituting back:

Therefore, \(\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}\).

We know the Maclaurin series for \(\sin(t)\):

Dividing by \(t\):

Integrating term by term:

Using the limits of integration:

Therefore, the power series representation for \(f(x) = \int_0^x \frac{\sin(t)}{t} \, dt\) is:

This function is known as the sine integral \(\text{Si}(x)\).