7 Series

A series is written as

    \[\sum_{n=1}^{\infty} a_n=a_1+a_2+....,\]

and the definitions of convergent and divergent series are stated by Larson and Edwards as follows [15].



Definition I.14

For the infinite series \sum_{n=1}^{\infty} a_n, the nth partial sum is 


      \[S_n=a_1+a_2+...+a_n.\]


If the sequence of partial sums \{S_n\} converges to S, then the series \sum_{n=1}^{\infty} a_n converges. The limit S is called the sum of the series. 


      \[S=a_1+a_2+...+a_n+...\]


 If \{S_n\} diverges, then the series diverges.


As with sequences, we are concerned with whether a series converges or diverges. There are some types of series that have their own set of rules for whether they converge or not, such as geometric series and p-series. If a series is not one of these special types, then we have tests that can be used to determine convergence, such as the Direct Comparison and Alternating Series tests.


Geometric Series


A geometric series is a special type of series. Not only is it easy to tell whether such a series converges or diverges, but it is also easy to find the sum that the series converges to. Let’s look at the following theorem and proof from Larson and Edwards [15].


Theorem I.15 (Convergence of a Geometric Series)


A geometric series with ratio r diverges when |r|\geq 1. If 0<|r|<1, then the series converges to the sum 


      \[\sum_{n=0}^{\infty} ar^n=a+ar+ar^2+...=\frac{a}{1-r}.\]




Proof.

It is easy to see that the series diverges when r=\pm 1. If r\neq \pm 1, then we have the partial sum


      \begin{align*} S_n&=a+ar+ar^2+...+ar^{n-1}\\ rS_n&=ar+ar^2+ar^3+...+ar^n \\ S_n-rS_n&=a-ar^n\\ S_n(1-r)&=a(1-r^n)\\ S_n&=\frac{a}{1-r}(1-r^n) \end{align*}


When 0<|r|<1, it follows that \displaystyle\lim_{n\rightarrow\infty}r^n=0. So,


      \[\lim_{n\rightarrow\infty} S_n=\lim_{n\rightarrow\infty}\left[\frac{a}{1-r}(1-r^n)\right]=\frac{a}{1-r}.\]


 Thus, the series converges, and its sum is \displaystyle\frac{a}{1-r}.


When r<-1, it follows that \lim_{n\rightarrow\infty}r^n alternates between -\infty and \infty, and when r>1, it follows that \displaystyle\lim_{n\rightarrow\infty}r^n=\infty. So if |r|>1, 


      \[\lim_{n\rightarrow\infty} S_n=\lim_{n\rightarrow\infty}\left[\frac{a}{1-r}(1-r^n)\right]=\pm\infty.\]


 Thus, the limit does not exist, and the series diverges.


Let’s now do an example I completed for Calculus II that comes from Larson and Edwards [15].




Example 15

Does the series \displaystyle\sum_{n=0}^{\infty} 5\left(\displaystyle\frac{2}{3}\right)^n converge or diverge? If it converges, what is the sum?Solution

We have that a=5 and r=\frac{2}{3}. Because

0<\left|\frac{2}{3}\right|<1, the series converges, and the sum is 


      \[\frac{5}{1-\frac{2}{3}}=\frac{5}{\frac{1}{3}}=5\cdot 3=15\]


 by Theorem I.15.


Another special type of series for which it is easy to determine convergence and divergence is the p-series.


P-series


Consider the following theorem provided by Larson and Edwards [15].


Theorem I.16


The p-series 


      \[\sum_{n=1}^{\infty} \frac{1}{n^p}=\frac{1}{1^p}+\frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+...\]


 converges for p>1 and diverges for 0<p\leq1.


Let’s now look at how to use this theorem. This problem is one I completed for Calculus II and comes from Larson and Edwards [15].






Example 16


Does 1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+... converge or diverge?


‌


Solution


      \begin{align*} 1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+...&=\sum_{n=1}^{\infty}\frac{1}{n\cdot n^{\frac{1}{2}}}\\ &=\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}\\ \end{align*}


Since p=\frac{3}{2} and \frac{3}{2}>1, this series converges by Theorem I.16.




Because many series do not fall into nice categories such as geometric series and p-series, we need tests that we can use to help us determine convergence and divergence.


Direct Comparison Test


The first test for convergence and divergence that we are going to discuss is the direct comparison test. This test is stated as a theorem by Larson and Edwards as follows [15].


Theorem I.17




Let 0<a_n\leq c_n for all n.


    

  1. If \displaystyle\sum_{n=1}^{\infty} c_n converges, then \displaystyle\sum_{n=1}^{\infty} a_n converges.
    2. 

  2. If \displaystyle\sum_{n=1}^{\infty} a_n diverges, then \displaystyle\sum_{n=1}^{\infty} c_n diverges.
    3. 





Proof.

To prove the first part, let \sum_{n=1}^{\infty}c_n=L and let 


      \[A_n=a_1+a_2+...+a_n\]


 and 


      \[C_n=c_1+c_2+...+c_n.\]


 Since \sum_{n=1}^{\infty}c_n=L, \{C_n\} converges to L by Definition I.14.  Additionally, because 0<a_n\leq c_n for all n, we have 


      \[C_n<C_{n+1}\leq L\]


 and 


      \[A_n\leq C_n\leq L.\]


 So, the sequence \{A_n\} is bounded above. Because a_n>0, \{A_n\} is also increasing and thus bounded below by A_1. By Theorem I.13, \{A_n\} converges, and \sum_{n=1}^{\infty} a_n converges by Definition I.14.


Therefore, if 0<a_n\leq c_n for all n and \sum_{n=1}^{\infty} c_n converges, \sum_{n=1}^{\infty} a_n converges as well. Because part 2 is the contrapositive statement of part 1, part 2 is also true. (See Part II: Chapter 11.)


This proof is one I completed as a challenge for myself with some help from the proof given for this test in Larson and Edwards [15]. Let’s now work through an example I completed for Calculus II that comes from Larson and Edwards [15].






Example 17


Using the Direct Comparison Test, does \displaystyle\sum_{n=0}^{\infty} \displaystyle\frac{4^n}{5^n+3} converge or diverge?


‌ ‌


Solution

The series 


      \[\sum_{n=0}^{\infty} \frac{4^n}{5^n+3}\]


 looks very similar to the geometric series 


      \[\sum_{n=0}^{\infty} \left(\frac{4}{5}\right)^n.\]


 We know that this geometric series converges by Theorem I.15 so the Direct Comparison test will help us only if for all n\geq 0, 


      \[\frac{4^n}{5^n+3}\leq \frac{4^n}{5^n}=\left(\frac{4}{5}\right)^n.\]


We can show this is true the following way.


      \begin{align*} 5^n+3&> 5^n>0\\ \frac{1}{5^n+3}&< \frac{1}{5^n}\\ \frac{4^n}{5^n+3}&< \frac{4^n}{5^n} \end{align*}


Because 0<\displaystyle\frac{4^n}{5^n+3}< \displaystyle\left(\frac{4}{5}\right)^n for all n\geq 0 and \displaystyle\sum_{n=0}^{\infty} \left(\frac{4}{5}\right)^n converges, \displaystyle\sum_{n=0}^{\infty} \frac{4^n}{5^n+3} converges by the Direct Comparison Test.




We are now going to discuss one other test, the alternating series test, before concluding our discussion of series.


Alternating Series Test


The alternating series test helps us determine if a series is converging. Larson and Edwards state the test as follows [15].


Theorem I.18 (Alternating Series Test)




Let a_n>0. The alternating series 


      \[\sum_{n=1}^{\infty}(-1)^na_n\,\,\text{and}\,\,\sum_{n=1}^{\infty}(-1)^{n+1}a_n\]


 converge when the two conditions listed below are met.


    

  1. \displaystyle\lim_{n\rightarrow\infty} a_n=0
    2. 

  2. a_{n+1}\leq {a_n}, for all n
    3. 





Proof.

Consider the alternating series \displaystyle\sum_{n=1}^{\infty}(-1)^{n}a_n. For this series, the partial sum (where 2n is even) 


      \[S_{2n}=(-a_1+a_2)+(-a_3+a_4)+(-a_5+a_6)+...+(-a_{2n-1}+a_{2n})\]


 has all nonpositive terms because 0<a_{n+1}\leq a_{n}, and therefore \{S_{2n}\} is a nonincreasing sequence. But the partial sum can be rewritten as 


      \[S_{2n}=-a_1+(a_2-a_3)+(a_4 -a_5)+...+(a_{2n-2}-a_{2n-1})+a_{2n}\]


 which implies that S_{2n}\geq -a_1 for every integer n (because 0<a_{n+1}\leq a_{n}). So, \{S_{2n}\} is a bounded, nonincreasing sequence that converges to some value L by Theorem I.13. Notice that S_{2n-1}=S_{2n}-a_{2n}. Because \displaystyle\lim_{n\rightarrow\infty} a_n=0, \displaystyle\lim_{n\rightarrow\infty} a_{2n}=0. So we have,


      \begin{align*} \displaystyle\lim_{n\rightarrow\infty} S_{2n-1}&=\lim_{n\rightarrow\infty} S_{2n}-\lim_{n\rightarrow\infty} a_{2n}\\ &=L-0\\ &=L \end{align*}


Because both S_{2n} and S_{2n-1} converge to the same limit L, it follows that \{S_n\} also converges to L. Consequently, the given alternating series converges by Definition I.14.


This proof is one I completed for Calculus II. The proof for when the series is 


      \[\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}a_n\]


 can be done in a similar manner and is found in Larson and Edwards [15]. Let’s now work through an example I completed for Calculus II that comes from Larson and Edwards [15].






Example 18


Does \displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\cos (n\pi) converge or diverge?


‌


Solution

Because 


      \[\sum_{n=1}^{\infty} \frac{1}{n}\cos (n\pi)=\sum_{n=1}^{\infty} \frac{1}{n}(-1)^n\]


 and \displaystyle\frac{1}{n}>0, we need to check the two conditions of the Alternating Series test to see if this test applies. Because


      \[\lim_{n\rightarrow\infty} \frac{1}{n}=\frac{1}{\infty}=0\]


the first condition holds. Additionally, because 


      \[0<n<n+1 \Rightarrow \frac{1}{n+1}<\frac{1}{n}\]


 for all n, the second condition holds. Therefore, \displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\cos (n\pi) converges by the Alternating Series Test.




Our final calculus topic we will discuss is differential equations. Differential equations contain derivatives within them, and in order to solve them, we must use integration. There is so much we could discuss, but we are going to settle with one specific type, the separable differential equation.





					
		
					

						

							
							

								
License

								
Portfolio for Bachelor of Science in Mathematics  Copyright © by Abigail E. Huettig. All Rights Reserved.
							

														

								
Share This Book