Limits, Sequences, Series: Learning Objective 4

To find the terms of the sequence, substitute the values $n=1, 2, 3, 4$ into the formula for $a_n$.

$\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots\}$

We calculate the first four terms by plugging in $n=1, 2, 3, 4$:

• $a_1 = \frac{1}{1+1} = \frac{1}{2}$
• $a_2 = \frac{2}{2+1} = \frac{2}{3}$
• $a_3 = \frac{3}{3+1} = \frac{3}{4}$
• $a_4 = \frac{4}{4+1} = \frac{4}{5}$

The sequence begins $\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots, \frac{n}{n+1}, \dots\}$.

To evaluate the limit of this rational expression at infinity, divide both the numerator and the denominator by the highest power of $n$.

The limit is $1$. This means that the terms of the sequence get arbitrarily close to 1 as $n$ becomes very large.

1. Divide by the highest power of n: The highest power of $n$ is $n^1$. $$ \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{n/n}{(n/n)+(1/n)} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} $$ 2. Evaluate the limit: As $n \to \infty$, the term $\frac{1}{n} \to 0$. $$ = \frac{1}{1+0} = 1 $$ Interpretation: The sequence is convergent, and it converges to 1. This means that as we go further and further in the sequence, the value of the terms gets closer and closer to 1.

A sequence converges if its limit as $n \to \infty$ is a finite number. Evaluate this limit by dividing the numerator and denominator by the highest effective power of $n$. In the denominator, treat $\sqrt{n}$ as the dominant term.

Divergent.

We evaluate the limit of the sequence as $n \to \infty$. $$ \lim_{n \to \infty} \frac{n}{\sqrt{10+n}} $$ Divide the numerator and denominator by the highest power of $n$ from the denominator, which is effectively $\sqrt{n}$. In the numerator, we divide by $\sqrt{n}$. $$ = \lim_{n \to \infty} \frac{n/\sqrt{n}}{\sqrt{10+n}/\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{\frac{10}{n}+\frac{n}{n}}} = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{\frac{10}{n}+1}} $$ As $n \to \infty$, the numerator $\sqrt{n} \to \infty$ and the denominator approaches $\sqrt{0+1} = 1$. $$ \to \frac{\infty}{1} = \infty $$ Since the limit is infinite, the sequence is divergent.

One method is to consider the associated function $f(x) = \frac{x}{x^2+1}$ and show that its derivative, $f'(x)$, is negative for all $x \ge 1$.

The sequence is shown to be decreasing by demonstrating that the derivative of the corresponding function $f(x) = \frac{x}{x^2+1}$ is negative for $x \ge 1$.

Let $f(x) = \frac{x}{x^2+1}$. We find its derivative using the quotient rule: $$ f'(x) = \frac{(x^2+1)\frac{d}{dx}(x) - x\frac{d}{dx}(x^2+1)}{(x^2+1)^2} $$ $$ = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} $$ A sequence is decreasing if $f'(x) < 0$. The denominator $(x^2+1)^2$ is always positive. So the sign of $f'(x)$ is determined by the numerator, $1-x^2$.

For $x > 1$, $x^2 > 1$, which means $1-x^2 < 0$.

Since $f'(x) < 0$ for all $x > 1$, the function $f(x)$ is decreasing for $x \ge 1$. This implies that the sequence $a_n = f(n)$ is decreasing for all $n \ge 1$.

The sum of an infinite series is, by definition, the limit of its sequence of partial sums as $n$ approaches infinity.

$\frac{2}{3}$

The sum of the series, $S$, is the limit of the sequence of its partial sums, $s_n$. $$ S = \lim_{n \to \infty} s_n = \lim_{n \to \infty} \frac{2n}{3n+5} $$ To evaluate this limit, we divide the numerator and denominator by the highest power of $n$, which is $n$: $$ S = \lim_{n \to \infty} \frac{2n/n}{(3n/n)+(5/n)} = \lim_{n \to \infty} \frac{2}{3+\frac{5}{n}} $$ As $n \to \infty$, the term $\frac{5}{n} \to 0$. $$ S = \frac{2}{3+0} = \frac{2}{3} $$

First, identify the first term, $a$, and the common ratio, $r$. If $|r|<1$, the series converges and its sum is given by the formula $S = \frac{a}{1-r}$.

$3$

1. Identify the first term and common ratio:

• The first term is $a=5$.
• The common ratio is $r = \frac{a_2}{a_1} = \frac{-10/3}{5} = -\frac{10}{15} = -\frac{2}{3}$.

2. Check for convergence: Since $|r| = |-\frac{2}{3}| = \frac{2}{3} < 1$, the geometric series converges.

3. Calculate the sum: $$ S = \frac{a}{1-r} = \frac{5}{1 - (-\frac{2}{3})} = \frac{5}{1+\frac{2}{3}} = \frac{5}{\frac{5}{3}} = 5 \cdot \frac{3}{5} = 3 $$

Rewrite the expression for the general term to see if it fits the form of a geometric series, $ar^{n-1}$ or $ar^n$. Check if the absolute value of the common ratio is less than 1.

Divergent.

Let's rewrite the term $a_n = 2^{2n}3^{1-n}$ to identify if it is a geometric series. $$ a_n = (2^2)^n \cdot 3^1 \cdot 3^{-n} = 4^n \cdot 3 \cdot \frac{1}{3^n} = 3 \left(\frac{4}{3}\right)^n $$ This is a geometric series with common ratio $r = \frac{4}{3}$.

A geometric series converges if and only if $|r| < 1$. In this case, $|r| = |\frac{4}{3}| = \frac{4}{3} > 1$.

Since the absolute value of the common ratio is greater than 1, the series is divergent.

Use the Direct Comparison Test. Find a simpler, larger series that is known to converge. A good choice for comparison is the geometric series $\sum_{n=1}^{\infty} \frac{1}{2^n}$.

Convergent.

We will use the Direct Comparison Test.

1. Find a series for comparison: Let $a_n = \frac{1}{2^n+1}$. Let's compare this to $b_n = \frac{1}{2^n}$.

2. Establish the inequality: For all $n \ge 1$, we know that $2^n+1 > 2^n$. Taking the reciprocal reverses the inequality sign: $$ \frac{1}{2^n+1} < \frac{1}{2^n} $$ So, $a_n < b_n$.

3. Check the convergence of the comparison series: The series $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{2^n}$ is a geometric series with common ratio $r=\frac{1}{2}$. Since $|r|<1$, this series converges.

4. Conclusion: By the Direct Comparison Test, since our series $\sum a_n$ is smaller than a known convergent series $\sum b_n$, our series must also converge.

Try the Integral Test. The associated function $f(x) = \frac{\ln(x)}{x}$ is positive, continuous, and decreasing for $x \ge e$. Evaluate the improper integral $\int_1^\infty f(x)dx$.

Divergent.

We use the Integral Test. Let $f(x) = \frac{\ln x}{x}$. The function is positive for $x > 1$ and is continuous. It can be shown to be decreasing for $x \ge e$. We can now evaluate the improper integral. $$ \int_1^\infty \frac{\ln x}{x} dx = \lim_{t \to \infty} \int_1^t \frac{\ln x}{x} dx $$ We use u-substitution: let $u=\ln x$, so $du = \frac{1}{x}dx$.
When $x=1, u=\ln(1)=0$. When $x=t, u=\ln(t)$. $$ = \lim_{t \to \infty} \int_0^{\ln t} u \, du = \lim_{t \to \infty} \left[ \frac{1}{2}u^2 \right]_0^{\ln t} = \lim_{t \to \infty} \frac{1}{2}(\ln t)^2 $$ As $t \to \infty$, $\ln t \to \infty$, so $(\ln t)^2 \to \infty$. $$ \frac{1}{2}\lim_{t \to \infty} (\ln t)^2 = \infty $$ Since the integral diverges, the series $\sum_{k=1}^{\infty} \frac{\ln(k)}{k}$ also diverges by the Integral Test.

Apply the Alternating Series Test. You need to check two conditions: (1) The magnitude of the terms, $b_n$, must be decreasing. (2) The limit of $b_n$ as $n \to \infty$ must be 0.

Convergent.

We use the Alternating Series Test on $\sum_{n=1}^\infty (-1)^{n-1}b_n$, where $b_n = \frac{1}{n}$.

We must check two conditions:
1. Is $b_n$ decreasing? We need to check if $b_{n+1} \le b_n$. $$ \frac{1}{n+1} \le \frac{1}{n} $$ This inequality is true for all $n \ge 1$ because the denominator $n+1$ is always greater than $n$. So, the terms are decreasing.

2. Is $\lim_{n \to \infty} b_n = 0$? $$ \lim_{n \to \infty} \frac{1}{n} = 0 $$ This condition is also met.

Since both conditions of the Alternating Series Test are satisfied, the series is convergent.