10-15-25 Series

Divergence tests

• Geometric Series

• Alternating Series tests

• Integral test

• Comparison

• P Series

• Absolute convergence test

Practice

Step 1: Check if the series is divergent:

• $\sum_{n=1}^{\infty}a_{n}$ if $lim_{n\rightarrow\infty}a_{n}\ne0$

Step 2: Is the series one of the popular series (Geometric series, p-series, alternating series)?

Geometric series: $\sum_{n=0}^\infty a r^n$

• Converges if -1 < r < 1

P-series: $\sum_{n=1}^\infty \frac 1{n^P}$

• Converges if p > 1

Alternating series: $\sum_{n=1}^\infty (-1)^n a_n$

• Converges if a_n > a_{n+1} > 0 and $lim_{n \rightarrow \infty} a_n = 0$

Step 3: Check if you can use the ordinary comparison test or the limit comparison test

The Comparison test

• Suppose $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are positive terms

  • If $\sum_{n=1}^\infty b_n$ converges and $a_n \le \ b_n$ for all n, then $\sum_{n=1}^\infty$ converges

  • If $\sum_{n=1}^\infty b_n$ diverges and $a_n \ge \ b_n$ for all n, then $\sum_{n=1}^\infty a_n$ diverges

Limit Comparison Test

• If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are positive-term series, and if $lim_{n \rightarrow \infty} \frac {a_n}{b_n} = K$ where K > 0 and finite, then either both series converge or both diverge

Step 4: Check if you can use the ratio test. For a series $\sum a_n$, let $L = lim_{n \rightarrow \infty} |\frac {a_{n+1}}{a_n}|$

• If L < 1, the series is absolutely convergent and thus convergent

• If L > 1, the series is divergent

• If $L = 1$ the test fails; no conclusion regarding convergence/divergence

Step 5: Check if you can use the absolute convergence test. (if the question asks about ABSOLUTE CONVERGENCE)

• If the series $\sum |a_n|$ is convergent, then $\sum a_n$ is absolutely convergent

Step 6: Check if you can use the integral test

• Let f be a function that is positive-valued, continuous and decreasing on $[1, \infty)$, and $a_n = f(n)$ for all integers n. Then

  • $\int_1 ^\infty f(x) dx$

  • $\sum_{n}^\infty f(n)$

• either both converge or diverges

Steps for determining whether the series $\sum_{n=1}^\infty a_n$ is convergent or not

1. Check if the series is divergent (Use the divergence test)

2. Is the series one of the popular series (Geometric series, p-series, alternating series)?

3. Check if you can use the ordinary comparison test or the limit comparison test.

4. Check if you can use the ratio test

5. Check if you can use the absolute convergence test (if the question asks about absolute convergence)

6. Check if you can use the integral test.

Q6. Choose the right answer. Consider the series

$\sum_{n=1}^\infty \frac {5^n * n}{1 * 3 * 5 …(2n + 1)}$

a) The series is absolutely convergent

b) The series is conditionally convergent

c) The series is divergent

Solution of Q6.

$a_n = \frac {5^n n!}{1 × 3 × 5 … (2n + 1)}$

Lets call the denominator $D_n = 1 × 3 × 5 … (2n + 1)$

Then

• $D_{n+1} = D_n * (2n+3)$

Compute

• $\frac {a_{n+1}}{a_n} = \frac {5^{n+1}(n+1)!/D_{n+1}}{5^{n}n!/D_n}$

Simplify

• $\frac {a_{n+1}}{a_n} = 5 × (n+1) × \frac {D_n}{D_{n+1}}$

and since $D_{n+1} = D_n (2n+3)$, this gives:

• $\frac {a_{n+1}}{a_n} = 5 × (n + 1) × \frac {1}{(2n+3)}$

• $L = lim_{n \rightarrow \infty} \frac {a_{n+1}}{a_n} = lim_{n \rightarrow \infty} 5 \frac {n + 1}{2n + 3}$

Divide numerator and denominator by n:

• If L < 1: Absolutely convergent

• If L > 1: Divergent

• If L = 1: Inconclusive

Here, L = 2.5 > 1

More practice:

$\sum_{n=1}^\infty \frac {(-2)^{2n}}{5^n}$

Can be solved using the geometric series:

$\sum_{n=1}^\infty \frac {(-2)^{2n}}{5^n} \Rightarrow \frac {4^n}{5^n} \Rightarrow (\frac 45)^n$

$\sum_{n=1}^\infty (\frac 45)^n$ is a geometric series ($\sum_{n=1}^\infty a_nr^n$)

A geometric series converges if the absolute value of its common ratio, $|r|$, is less than 1. The common ratio for this series is $r =\frac 45$

Since r < 1, the series converges

Some more practice:

$\sum_{n=1}^\infty (-1)^n \frac {n+2}{4n+5}$

This is an alternate series ($\sum_{n=1}^\infty (-1)^na_n$):

• The alternate series converges if a_n > a_{n+1} > 0 and $lim_{n \Rightarrow \infty} a_n = 0$

$a_n = \frac {n+2}{4n+5}$

$a_1 \Rightarrow \frac {1+2}{4(1) + 5} = \frac 39 \Rightarrow 0. \overline {333}$

$a_2 \Rightarrow \frac {2+2}{4(2) + 5} = \frac 4{13} \Rightarrow 0.308$

$a_3 \Rightarrow \frac {3+2}{4(3)+5} = \frac 5{17} \Rightarrow 0.294$

a_1 > a_2 > a_3 > 0

$lim_{n \Rightarrow \infty}a_n = lim_{n \Rightarrow \infty} \frac {n + 2}{4n + 5}$

• To find the limit, divide both the numerator and the denominator by the highest power of n, which is n

$lim_{n \Rightarrow \infty} \frac {\frac nn + \frac 2n}{\frac {4n}n + \frac 5n} \Rightarrow \frac {1 + \frac 2n}{4 + \frac 5n}$

• As $n \Rightarrow \infty$, the terms $\frac 2n$and$\frac 5n$ approach 0. Therefore, the limit is $\frac 14$

Since a_n > a_{n+1} > a_{n+2} + … > 0 m_{n \Rightarrow \infty} a_n = 0$, <strong><u>but</u></strong>$lim_{n\Rightarrow \infty} \ne 0$the series diverges</p><p></p><p><strong><u>Even more practice!:</u></strong></p><p>$\sum_{n=1}^\infty \frac {n^4}{n^5 + 5}$</p><p></p><p>We can find if the series converges or diverges using the Limit Comparison test</p><p></p><p>$a_n = \frac {n^4}{n^5 + 5}$</p><p>$lim_{n \Rightarrow \infty} \frac {n^4}{n^5 + 5} \Rightarrow lim_{n \Rightarrow \infty} \frac {n^4}{n^5} = \frac 1n$</p><p></p><ul><li><p>Therefore, we will compare the given series to the harmonic series$\sum_{n=1}^\infty b_n$where$b_n = \frac 1n$</p></li></ul><p></p><p>The limit comparison test states that if we have two series$\sum_{n=1}^\infty a_n$and$\sum_{n=1}^\infty b_n$with positive terms, and if the limit$L = lim_{n \Rightarrow \infty} \frac {a_n}{b_n}$is a finite positive number$L > 0$, then both series either converge or diverge.</p><p></p><p>$\frac {a_n}{b_n} \Rightarrow \frac {\frac {n^4}{n^5 + 5}}{\frac 1n} \Rightarrow \frac {n^4}{n^5 + 5} × \frac n1 \Rightarrow \frac {n^5}{n^5 + 5}$</p><p>$L = lim_{n \Rightarrow \infty} \frac{n^5}{n^5 + 5} = \frac {n^5}{n^5} = 1$</p><p>$L > 0$</p><p></p><p>Concluding convergence or divergence:</p><p>$\sum_{n=1}^\infty \frac 1n$is a harmonic series, which is a p-series with p = 1$\sum_{n=1}^\infty \frac 1{n^p}$diverges if$p \le \ 1$. Since$p = 1$, the harmonic series diverges.</p><p>By the limit comparison test, because the harmonic series diverges,$\sum_{n=1}^\infty \frac {n^4}{n^5 + 5}$also diverges </p><p></p><p><strong><u>THE FINAL PRACTICE:</u></strong></p><p>$\sum_{n=1}^\infty \frac {sin 2 n}{n³}$</p><p></p><p>Comparison test using p-series</p><p></p><p>$a_n = |\frac {sin(2n)}{n³}|$</p><p>$sin(x) \le \ 1 \Rightarrow |\frac {sin(2n)}{n³}| \le \ \frac 1{n³}$</p><p>$b_n = \frac{1}{n³}$</p><p>$\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty \frac {1}{n³} is a p-series with p = 3. Since p > 1, this series converges.

Because a_n \le b_n$and$\sum b_n$converges, the series$\sum a_n = \sum_{n=1}^\infty |\frac {sin(2n)}{n³}|$$ converges by the comparison test