## 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

(Click here for the limit step by step procedures)

=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