Thursday, March 26, 2020

Advanced Math Solutions - Series Convergence Calculator, Series Ratio Test

In our Series blogs, we’ve gone over four types of series, Geometric, p, Alternating, and Telescoping, and their convergence tests. Now, we will focus on convergence tests for any type of infinite series, as long as they meet the tests’ criteria. We will start off with the Series Ratio Test.

What does the Series Ratio Test do?

The goal of the Series Ratio Test is to determine if the series converges or diverges by evaluating the ratio of the general term of the series to its following term. The test determines if the ratio absolutely converges. A series absolutely convergences if the sum of the absolute value of the terms is finite. If there is absolute convergence, then there is convergence. This will make more sense, once you see the test and try out a few examples.

Some caveats: The test will not determine what the series will converge to. The test may also result in inconclusive results.

What is the Series Ratio Test?

Given a series, ∑a_n , determine L such that

lim_{n→∞} \mid\frac{a_{n+1}}{a_n}\mid=L,where a_n≠0
If L<1, then ∑a_n  converges.
If L>1, then ∑a_n  diverges.
If L=1, then the test is inconclusive.

The test may seem pretty straight forward and simple, but determining what type of series to use this test on is not.

What types of series should the test be used on?

While there is no straight forward answer to this question, this test is typically helpful when determining convergence for series with exponential functions or factorials.

Now that we know what the series ratio test is, let’s see some examples of how it is used.

∑_{n=0}^∞\frac{2^n}{n!}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

lim_{n→∞} \mid\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}\mid=lim_{n→∞} \mid\frac{2^{n+1}∙n!}{(n+1)!∙2^n }\mid
=lim_{n→∞} \mid\frac{2∙n!}{(n+1)!}\mid

(Note: 2^{n+1} = 2^n ∙ 2^1 )

=lim_{n→∞} \mid\frac{2}{(n+1) }\mid

(Note: \frac{n!}{(n+m)!}=\frac{1}{(n+1)∙(n+2)  ∙ ∙ ∙(n+m) })

=0=L

For detailed procedures on how to determine the limit, please see the step by step procedures in the link.

2. Given L, determine convergence

Since L = 0 and is < 1, the series converges.

∑_{n=1}^∞\frac{6^n}{n}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

lim_{n→∞} \mid\frac{\frac{6^{n+1}}{n+1}}{\frac{6^n}{n}}\mid=lim_{n→∞} \mid\frac{6^{n+1}∙n}{(n+1)∙6^n }\mid

=lim_{n→∞} \mid\frac{6n}{(n+1) }\mid

=6∙lim_{n→∞} \mid\frac{1}{(1+\frac{1}{n}) }\mid

(Note: \frac{n}{n+1 }=  \frac{1}{1+\frac{1}{n}})

=6∙1=6=L

2. Given L, determine convergence

Since L = 6 and is > 1, the series diverges.

∑_{n=0}^∞\frac{2^2n}{3^2n}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

lim_{n→∞} \mid\frac{\frac{2^{2(n+1)}}{3^{2(n+1)}} }{\frac{2^2n}{3^2n} }\mid=lim_{n→∞} \mid\frac{(\frac{2}{3})^{2(n+1)}}{(\frac{2}{3})^2n} \mid

=lim_{n→∞} \mid(\frac{2}{3})^{2(n+1)-2n} \mid

=lim_{n→∞} \mid(\frac{2}{3})^2 \mid

=\frac{4}{9}=L

2. Given L, determine convergence

Since L = \frac{4}{9} and is < 1, the series converges.

Last example:
∑_{n=1}^∞\frac{1}{n}

1. Determine the limit of lim_{n→∞} \mid\frac{a_{n+1}}{a_n} \mid

lim_{n→∞} \mid\frac{\frac{1}{n+1}}{\frac{1}{n}}\mid=lim_{n→∞} \mid\frac{n}{n+1}\mid

=lim_{n→∞} \mid\frac{1}{1+\frac{1}{n}}\mid

=1=L

2. Given L, determine convergence

Since L = 1, the test is inconclusive.

You may be wondering what to do if the ratio series test is inconclusive. In the next couple of blog posts, we will be discussing other convergence tests that can be used when the ratio test is inconclusive. For more practice on the Ratio Series Test, check out Symbolab’s Practice.

Until next time,

Leah

Wednesday, March 18, 2020

Advanced Math - Series Convergence Calculator, Telescoping Series Test

Last blog post, we went over what an alternating series is and how to determine if it converges using the alternating series test. In this blog post, we will discuss another infinite series, the telescoping series, and how to determine if it converges using the telescoping series test.

If it isn’t clear right away, telescoping is synonymous with the word collapsing. A telescoping series is a series where almost all the terms cancel with the preceding or following term leaving just the initial and final terms, i.e. a series that can be collapsed into a few terms.

Let’s see what this looks like . . .

∑_{n=1}^∞\frac{1}{n(n+1)}= ∑_{n=1}^∞\frac{1}{n}-\frac{1}{n+1}

= (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+ ...+(\frac{1}{n}-\frac{1}{n+1})

=1-\frac{1}{n+1}

As you can see, we are able to cancel out all terms except the first and last.

Now that we’ve discussed what a telescoping series is, let’s go over the telescoping series test.

Telescoping Series Test:

For a finite upper boundary, ∑_{n=k}^N(a_{n+1}-a_n )=a_{N+1 }-a_k
For an infinite upper boundary, if a_n→0*, then ∑_{n=k}^∞(a_{n+1}-a_n )= -a_k
*If a_n doesn’t converge to 0, then the series diverges.

In regards to infinite series, we will focus on the infinite upper boundary scenario. In order to use this test, you will need to manipulate the series formula to equal a_{n+1}-a_n where you can easily identify what a_{n+1} and a_n are. Also, please note that if you are able to manipulate the series in this form, you can confirm that you have a telescoping series. With practice, this will come more naturally.

Let’s see some examples to better understand.

∑_{n=1}^∞\frac{5}{n}-\frac{5}{n+1}

1. Convert the series into the form a_{n+1}-a_n

\frac{5}{n}-\frac{5}{n+1}= -\frac{5}{n+1}-(-\frac{5}{n})

a_{n+1}=-\frac{5}{n+1}

a_n=-\frac{5}{n}

2. Determine if a_n→0

a_n=-\frac{5}{n}= -5(\frac{1}{n})

Since \frac{1}{n} converges to 0, -\frac{5}{n} converges to 0.

3. Calculate -a_k

k=1

a_n=-\frac{5}{n}

-a_k=-(-\frac{5}{1})=5
The series converges to 5.

∑_{n=1}^∞\frac{6}{(n+1)(n+2)}

1. Convert the series into the form a_{n+1}-a_n

∑_{n=1}^∞\frac{6}{(n+1)(n+2)}= 6∙∑_{n=1}^∞\frac{1}{(n+1)(n+2)}

\frac{1}{(n+1)(n+2)}= -(\frac{1}{n+2})-(-\frac{1}{n+1})

a_{n+1}=-\frac{1}{n+2}

a_n=-\frac{1}{n+1}

2. Determine if a_n→0

a_n=-\frac{1}{n+1}

-\frac{1}{n+1}  →0

3. Calculate -a_k

k=1

a_n=-\frac{1}{n+1}

-a_k=-(-\frac{1}{1+1})=\frac{1}{2}

6∙∑_{n=1}^∞\frac{1}{(n+1)(n+2)} =6∙\frac{1}{2}=3

The series converges to 3.

∑_{n=1}^∞\frac{1}{4n^2-1}

1. Convert the series into the form a_{n+1}-a_n

\frac{1}{4n^2-1}=-(\frac{1}{2(2n+1)} )-(-\frac{1}{2(2n-1)})

a_{n+1}= -(\frac{1}{2(2n+1)} )

a_n=-\frac{1}{2(2n-1)}

2. Determine if a_n→0

a_n=-\frac{1}{2(2n-1)} =-\frac{1}{4n-1}

-\frac{1}{4n-1}  →0

3. Calculate -a_k

k=1

a_n=-\frac{1}{2(2n-1)}

-a_k=-(-\frac{1}{2(2∙1-1)} )=\frac{1}{2}

The series converges to \frac{1}{2}.

The trickiest part of this is manipulating the series formula into a_{n+1}-a_n. Once you’re able to do this, the rest should be pretty simple. The key thing to remember about a telescoping series is that all the terms will cancel out, except the first and last term.

For more help on telescoping series, check out Symbolab’s Practice. Next blog post, I’ll go over the convergence test for a radio series.

Until next time,

Leah

Monday, October 21, 2019

Advanced Math Solutions - Series Convergence Calculator, Alternating Series Test

Last blog post, we discussed how to determine if an infinite p-series converges using the p-series test. In this blog post, we will discuss how to determine if an infinite alternating series converges using the alternating series test.

An alternating series is a series in the form ∑_{n=0}^∞(-1)^n∙a_n or ∑_{n=0}^∞(-1)^{n-1}∙a_n, where a_n>0 for all n. As you can see, the alternating series got its name from its terms that alternate between positive and negative values.

In some problems, you may come across alternating sign expressions that aren’t listed above. These various sign expressions can be simplified to equal the ones listed above. For example:

(-1)^{n+1}=(-1)^{n-1}∙(-1)^2=(-1)^{n-1}∙1=(-1)^{n-1}

Now that we know what an alternating series is, let’s discuss how to determine if the series converges, using the alternating series test.

Alternating Series Test:

An alternating series converges if all of the following conditions are met:
1. a_n>0 for all n
• a_n is positive
2. a_n>a_(n+1)  for all n≥N,where N is some integer
• a_n is always decreasing
3. lim_{n→∞} a_n=0

If an alternating series fails to meet one of the conditions, it doesn’t mean the series diverges. There are other tests that can be used to determine divergence.

Let’s see some examples of how to use the alternating series test to help you better understand.

∑_{n=1}^∞\frac{(-1)^{n+1}}{\sqrt{n+1}}
1. Is a_n>0 for all n

∑_{n=1}^∞\frac{(-1)^{n+1}}{\sqrt{n+1}}=∑_{n=1}^∞(-1)^{n+1}∙\frac{1}{\sqrt{n+1}}

a_n=\frac{1}{\sqrt{n+1}}>0 for all n

2. Is a_n>a_(n+1)  for all n≥N,where N is some integer?

∑_{n=1}^∞\frac{1}{\sqrt{n+1}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+ ...
\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}  and  \frac{1}{\sqrt{3}}>\frac{1}{\sqrt{4}}  and so on

a_n>a_(n+1)  for all n≥1

3. Is lim_{n→∞} a_n=0?

Let’s use this property to determine the limit of a_n:

lim_{n→∞}  \frac{1}{\sqrt{n+1}}=\frac{lim_{n→∞} 1}{lim_{n→∞} \sqrt{n+1}}=\frac{1}{∞}=0

4. Are all three conditions met?

Yes, the series converges.

∑_{n=1}^∞\frac{(-1)^n (n)}{n^2+2}

1. Is a_n>0 for all n?

a_n=\frac{n}{n^2+2}>0 for all n

2. Is a_n>a_{n+1}  for all n≥N,where N is some integer?

∑_{n=1}^∞\frac{n}{n^2+2}=\frac{1}{3}+\frac{2}{6}+\frac{3}{11}+\frac{4}{18}  ...=\frac{1}{3}+\frac{1}{3}+\frac{3}{11}+\frac{2}{9}  ...

\frac{1}{3}=\frac{1}{3},   \frac{1}{3}>\frac{3}{11},   \frac{3}{11}>\frac{2}{9}, and so on

a_n>a_(n+1)  for all n≥2

3. Is lim_{n→∞} a_n=0?

lim_{n→∞}  \frac{n}{n^2+1}=lim_{n→∞}  \frac{\frac{1}{n}}{1+\frac{1}{n^2}}=\frac{lim_{n→∞}  \frac{1}{n}}{lim_{n→∞} 1+\frac{1}{n^2}}=  \frac{0}{1+0}=0

4. Are all three conditions met?

Yes, the series converges.

Last example (:

∑_{n=2}^∞\frac{(-1)^{n+2}}{ln⁡(n)}

As discussed in the beginning, we can see a different alternating sign expression that wasn’t listed. However, we can simplify this expression to equal one that was listed.

(-1)^{n+2}=(-1)^n*(-1)^2=(-1)^n

1. Is a_n>0 for all n?

a_n=\frac{1}{ln⁡(n)}>0 for all n

2. Is a_n>a_{n+1}  for all n≥N,where N is some integer?

∑_{n=2}^∞\frac{1}{ln⁡(n)}=\frac{1}{ln⁡(2)}+\frac{1}{ln⁡(3)}+\frac{1}{ln⁡(4)}  ...

\frac{1}{ln⁡(2)}>\frac{1}{ln⁡(3)},   \frac{1}{ln⁡(3)}>\frac{1}{ln⁡(4)},    and so on

a_n>a_{n+1}  for all n≥2

3. Is lim_{n→∞} a_n=0?

lim_{n→∞}  \frac{1}{ln⁡(n)}=\frac{lim_{n→∞} 1}{lim_{n→∞} ln⁡(n)}=\frac{1}{∞}=0

4. Are all three conditions met?

Yes, the series converges.

The alternating series test has a lot of parts to it, but as long as you remember the three conditions, you’ll be able to master this topic! For more help or practice on the alternating series test, check out Symbolab’s Practice. Next blog post, I’ll go over the convergence test for telescoping series.

Until next time,

Leah

Wednesday, July 24, 2019

High School Math Solutions - Series Convergence Calculator, p-Series Test

Last blog post, we discussed what an infinite series is and how to determine if an infinite series converges using the geometric series test. In this blog post, we will discuss how to determine if an infinite series converges using the p-series test.

A p-series is a series of the form∑_{n=1}^∞\frac{1}{n^p} , where p is a constant power.

Here is an example of a p-series:

1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+ ...=\frac{1}{1^2} +\frac{1}{2^2} +\frac{1}{3^2} +\frac{1}{4^2} + ...=∑_{n=1}^∞\frac{1}{n^2}

So, how do we determine if the sum of a p-series converges to a finite number or diverges to an infinite number? We use the p-series test!

The following is the p-series test:

If the series is of the form ∑_{n=1}^∞\frac{1}{n^p}   , where p>0, then
If p>1, then the series converges.
If 0≤p<1, then the series diverges.

Unlike the geometric test, we are only able to determine whether the series diverges or converges and not what the series converges to, if it converges.

The p-series test is fairly simple, useful, and easy to remember.

Let’s see some examples of how to use it.

∑_{n=1}^∞\frac{1}{\sqrt{n}}
1. Determine the value of p

∑_{n=1}^∞\frac{1}{\sqrt{n}}= ∑_{n=1}^∞\frac{1}{n^{\frac{1}{2}}}

p=  \frac{1}{2}

2. Determine whether the series converges or diverges

Since p=  \frac{1}{2} and therefore 0≤p<1, the series diverges.

∑_{n=1}^∞\frac{n^2}{n^6}

1. Determine the value of p

∑_{n=1}^∞\frac{n^2}{n^6}   = ∑_{n=1}^∞\frac{1}{n^{6-2}} = ∑_{n=1}^∞\frac{1}{n^4}

p=4

In this step, I used the following exponent rule: \frac{x^a}{x^b} =\frac{1}{x^{b-a}}

2. Determine whether the series converges or diverges

Since p=4 and therefore p>1, the series converges.

∑_{n=1}^∞\frac{cos^2(n)+sin^2(n)}{n^2}

1. Determine the value of p

∑_{n=1}^∞\frac{cos^2(n)+sin^2(n)}{n^2} = ∑_{n=1}^∞\frac{1}{n^2}

p=2

In this step, I used the following trigonometric identity: sin^2(x)+cos^2(x)=1

2. Determine whether the series converges or diverges

Since p=4 and therefore p>1, the series converges.

The p-series test is pretty straightforward, helpful, and not too difficult. For more help or practice on the p-series test, check out Symbolab’s Practice. Next blog post, I’ll go over the convergence test for alternating series.

Until next time,

Leah

Monday, May 13, 2019

Advanced Math Solutions - Series Convergence Calculator, Geometric Series

Series are an important part of Calculus. In this next series of blog posts, I will be discussing infinite series and how to determine if they converge or diverge.

For a refresher:

A series is the sum of a list of terms that are generated with a pattern. A series is denoted with a summation symbol. An infinite series is a series that has an infinite number of terms being added together.

Here is an example of an infinite series:

With infinite series, it can be hard to determine if the series converges or diverges. Luckily, there are convergence tests to help us determine this!

In this blog post, I will go over the convergence test for geometric series, a type of infinite series.

A geometric series is a series that has a constant ratio between successive terms. A visualization of this will help you better understand.

Here’s a geometric series:

In this series, each following term is the product of the prior term and ⅓.

We can rewrite this geometric series using the summation notation.

In order to determine if a geometric series diverges or converges, you’ll need to follow and remember the following test/rule:

If the series is of the form ,

if |r|<1, then the geometric series converges to
if |r|≥1, then the geometric series diverges

Let’s see some examples to better understand.

1. Reference the geometric series convergence test

2. Determine the value of r

3. Determine if the series converges or diverges

The geometric series converges to \frac{5}{4}.

1. Reference the geometric series convergence test

2. Determine the value of r

3. Determine if the series converges or diverges

The geometric series diverges.

1. Reference the geometric series convergence test

2. Determine the value of r

3. Determine if the series converges or diverges

The geometric series converges to 6.

As you can see, it is not too difficult to determine if a geometric series converges or not. After doing some practice problems, you’ll get the hang of it very quickly. For more help or practice on geometric series, check out Symbolab’s Practice. Next blog post, I’ll go over the convergence test for p-series.

Until next time,

Leah

Monday, March 25, 2019

Advanced Math Solutions - Matrix Rank Calculator, Matrices

In the last two blog posts, we talked about Row Echelon Form (REF) and Reduced Row Echelon Form (RREF). In this blog post, we will talk about matrix rank. Determining a matrix’s rank will involve using REF or RREF, so make sure to review those blog posts before continuing on.

The rank of matrix is the dimension of the vector space created by its columns or rows. It is important to note that column rank and row rank are the same thing. We will find the rank of the matrix, by using the row rank.

Another way to think of this is that the rank of a matrix is the number of linearly independent rows or columns. Linearly independent means that no rows or columns can be the combination of the other rows or columns.

For example:

Here, Row 2 is a combination of Row 1 and Row 3 (Row 1 + Row 3). Therefore the rows are not linearly independent.

In order to determine the rank of a matrix:
1. Put the matrix in REF or RREF
2. Count the number of non-zero rows
Let’s see some examples. Please note that I won’t be going over how to put the matrices in REF or RREF.

1. Put the matrix in REF or RREF
The matrix is in RREF.
2. Count the number of non-zero rows
There are 3 non-zero rows. The rank of this matrix is 3.

1. Put the matrix in REF or RREF
The matrix is in REF.
2. Count the number of non-zero rows
There are 3 non-zero rows. The rank of this matrix is 3.

1. Put the matrix in REF or RREF
The matrix is in RREF.
2. Count the number of non-zero rows
There are 4 non-zero rows. The rank of this matrix is 4.

As you can see, finding the rank of a matrix is not hard. You just have to make sure you’ve mastered putting matrices in REF and RREF.

For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

Sunday, March 3, 2019

High School Math Solutions - Matrix Inverse Calculator, Matrices (Part 2)

In the last two blog posts, I talked about how to find the inverse of a matrix and how to calculate the determinant of the matrix. Please review these two blog posts before continuing, if you are not familiar with either topic.

As you saw in the inverse blog post, calculating the inverse of a matrix can require a lot of steps and some time. In this blog post, I will go over a shortcut for calculating the inverse of a 2x2 matrix.

Here are the steps for calculating the inverse of a 2x2 matrix, using the shortcut:

1. Calculate the determinant of matrix A
Reminder:

2. Reorganize matrix A

3. Multiply matrix A by \frac{1}{det(A)}

These steps can be summarized by this formula:

Not too difficult, right? Let’s see some examples.

1. Calculate the determinant of the matrix

2. Reorganize the matrix

3. Multiple the matrix by \frac{1}{det(A)}

1. Calculate the determinant of the matrix

2. Reorganize the matrix

3. Multiple the matrix by \frac{1}{det(A)}

1. Calculate the determinant of the matrix

2. Reorganize the matrix

3. Multiple the matrix by \frac{1}{det(A)}

This shortcut will help make calculating the inverse of a 2x2 matrix easier. That concludes our blog post series on matrices! For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah