Sequences and Series(page 1 of 5)Sections: Terminology and notation,Basic examples,Arithmetic and geometric sequences,Arithmetic series,Finite and infinite geometric series

A "sequence" (or "progression", in British English) is an ordered list of numbers; the numbers in this ordered list are called "elements" or "terms". A "series" is the value you get when you add up all the terms of a sequence; this value is called the "sum". For instance, "1, 2, 3, 4" is a sequence, with terms "1", "2", "3", and "4"; the corresponding series is the sum "1 + 2 + 3 + 4", and the value of the series is10.A sequence may be named or referred to as "A" or "An". The terms of a sequence are usually named something like "ai" or "an", with the subscripted letter "i" or "n" being the "index" or counter. So the second term of a sequnce might be named "a2" (pronounced "ay-sub-two"), and "a12" would designate the twelfth term.Note: Sometimes sequences start with an index ofn= 0, so the first term is actuallya0. Then the second term would bea1. The first listed term in such a case would be called the "zero-eth" term. This method of numbering the terms is used, for example, in Javascript arrays. Don't assume that every sequence and series will start with an index ofn= 1.A sequenceAwith termsanmay also be referred to as "{an}", but contrary to what you may have learned in other contexts, this "set" is actually an ordered list, not an unordered collection of elements. (Your book may use some notation other than what I'm showing here. Unfortunately, notation doesn't yet seem to have been entirely standardized for this topic. Just try always to make sure, whatever resource you're using, that you are clear on the definitions of that resource's terms and symbols.)ADVERTISEMENT

To indicate a series, we use either the Latin capital letter "S" or else the Greek letter corresponding to the capital "S", which is called "sigma" (SIGG-muh):

To show the summation of, say, the first through tenth terms of a sequence{an}, we would write the following:

The "n= 1" is the "lower index", telling us that "n" is the counter and that the counter starts at "1"; the "10" is the "upper index", telling us thata10will be the last term added in this series; "an" stands for the terms that we'll be adding. The whole thing is pronounced as "the sum, from n equals one to ten, of a-sub-n". The summation symbol above means the following:

a1+a2+a3+a4+a5+a6+a7+a8+a9+a10The written-out form above is called the "expanded" form of the series, in contrast with the more compact "sigma" notation.Copyright Elizabeth Stapel 2006-2011 All Rights ReservedAny letter can be used for the index, buti,j,k,andnare probably used more than any other letters.Sequences and series are most useful when there is a formula for their terms. For instance, if the formula foranis "2n+ 3", then you can find the value of any term by plugging the value ofninto the formula. For instance,a8= 2(8) + 3 = 16 + 3 = 19. In words, "an= 2n+ 3" can be read as "then-th term is given by two-enn plus three". The word "n-th" is pronounced "ENN-eth", and just means "the generic terman, where I haven't yet specified the value ofn."Of course, there doesn'thaveto be a formula for then-th term of a sequence. The values of the terms can be utterly random, having no relationship betweennand the value ofan. But sequences with random terms are hard to work with and are less useful in general, so you're not likely to see many of them in your classes.

Sequences and Series: Basic Examples(page 2 of 5)Sections:Terminology and notation, Basic examples,Arithmetic and geometric sequences,Arithmetic series,Finite and infinite geometric series

LetAn= {1, 3, 5, 7, 9}.

What is the value ofa3?

Find the value of The index ofa3isn= 3, so they're asking me for the third term, which is "5". The "value" they're asking for is the total, the sum, of all the termsanfroma1toa5; in other words:a1+a2+a3+a4+a5= 1 + 3 + 5 + 7 + 9 = 25value ofa3:5value of sum:25 Expand the following series and find the sum:

To find each term, I'll plug the value ofninto the formula. In this case, I'll be starting withn= 0and ending withn= 4.2(0) + 2(1) + 2(2) + 2(3) + 2(4) = 0 + 2 + 4 + 6 + 8 =20 List the first four terms of the sequence{an} = {n2}, starting withn= 1.I'll just plugninto the formula, and simplify:{a1,a2,a3,a4} = {12, 22, 32, 42} ={1, 4, 9, 16} List the first four terms of the following sequence, beginning withn= 0:

Many sequences and series containfactorials, and this is one of them. I'll evaluate in the usual way:

So the terms are:Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

Notice how, in that last example above, raising the1to the powernmade the signs alternate. This alternating pattern of signs crops up a lot, especially in calculus, so try to keep this "raising1to the powern" trick in mind. Find the sum of the first six terms ofAn, wherean= 2an1+an2,a1= 1, anda2= 1.This formula looks much worse than it really is; I just have to give myself some time, and dissect the formula carefully.They gave me the values of the first two terms, and then they gave me a formula that says that each term (after the first two terms) is a sum formed from the previous two terms. Plugging into the formula, I get:a3= 2a31+a32= 2a2+a1= 2(1) + (1) = 2 + 1 = 3a4= 2a41+a42= 2a3+a2= 2(3) + (1) = 6 + 1 = 7a5= 2a51+a52= 2a4+a3= 2(7) + (3) = 14 + 3 = 17a6= 2a61+a62= 2a5+a4= 2(17) + (7) = 34 + 7 = 41Now that I've found the values of the third through the sixth terms, I can find the value of the series; the sum is:1 + 1 + 3 + 7 + 17 + 41 =70 Write the following series using summation notation, beginning withn= 1:2 4 + 6 8 + 10The first thing I have to do isfigure out a relationshipbetweennand the terms in the summation. This series is pretty easy, though: each termanis twicen, so there is clearly a "2n" in the formula. I also have the alternating sign. If I use(1)n, I'll get2, 4, 6, 8, 10, which is backwards (on the signs) from what I want. But I can switch the signs by throwing in one more factor of1:(1)(1)n= (1)1(1)n= (1)n+1So the formula for then-th term isan= (1)n+1(2n). Sincenstarts at1and there are five terms, then the summation is:

Write the following using summation notation:

The only thing that changes from one term to the next is one of the numbers in the denominator. (If I "simplify" these fractions, I'll lose this information. Any time the terms of my sequence or series look oddly lumpy, I tend not to simplify those terms: that odd lumpiness almost certainly contains a hint of the pattern I need to find.) The changing numbers, as a list, are6, 7, and8. This looks like counting, but starting with6instead of1. Without any information to the contrary, I'll assume that this is the pattern.ADVERTISEMENT

But I need to relate these "counting" values to the counter, the index,n. Forn= 1, the number is6, orn+ 5. Forn= 2, the number is7, which is alson+ 5. Checking the pattern forn= 3, 3 + 5 = 8, which is the third number. Then the terms seems to be in the following pattern:

But how many terms are in the summation? The ellipsis (the "..." or "dot, dot, dot"in the middle) means that terms were omitted. However, now that I have the general pattern for the series terms, I can solve for the counter (the value ofn) in the last term:31 =n+ 531 5 =n+ 5 5n= 26

This tells me that there are26terms in this summation, so the series, in summation notation, is:

If the fractions (above) had been simplified and reduced, it would have been a lot harder to figure out a pattern. Unless the sequence is very simple or is presented in a very straightforward manner, it is possible that you won't be able to find a pattern, or might find a "wrong" pattern. Don't let this bother you terribly much: the "right" pattern is just the one that the author had in mind when he wrote the problem. Your pattern would be "wrong" only in that it is unexpected. But if you can present your work sensibly and mathematically, you should be able to talk your way into getting at least partial credit for your answer.Once you've learned the basic notation and terminology, you should quickly move on to the two common and straightforward sequence types....Arithmetic and Geometric Sequences(page 3 of 5)Sections:Terminology and notation,Basic examples, Arithmetic and geometric sequences,Arithmetic series,Finite and infinite geometric series

The two simplest sequences to work with are arithmetic and geometric sequences. An arithmetic sequence goes from one term to the next by always adding (or subtracting) the same value. For instance,2, 5, 8, 11, 14,...and7, 3, 1, 5,...are arithmetic, since you add3and subtract4, respectively, at each step. A geometric sequence goes from one term to the next by always multiplying (or dividing) by the same value. So1, 2, 4, 8, 16,...and81, 27, 9, 3, 1, 1/3,...are geometric, since you multiply by2and divide by3, respectively, at each step.The number added (or subtracted) at each stage of an arithmetic sequence is called the "common difference"d, because if you subtract (find the difference of) successive terms, you'll always get this common value. The number multiplied (or divided) at each stage of a geometric sequence is called the "common ratio"r, because if you divide (find the ratio of) successive terms, you'll always get this common value.Copyright Elizabeth Stapel 2006-2011 All Rights Reserved Find the common difference and the next term of the following sequence:3, 11, 19, 27, 35,...To find the common difference, I have to subtract a pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other:11 3 = 819 11 = 827 19 = 835 27 = 8The difference is always8, sod= 8. Then the next term is35 + 8 =43. Find the common ratio and the seventh term of the following sequence:ADVERTISEMENT

2/9, 2/3, 2, 6, 18,...To find the common ratio, I have to divide a pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other:

The ratio is always3, sor= 3. Then the sixth term is(18)(3) = 54and the seventh term is(54)(3) =162.

Since arithmetic and geometric sequences are so nice and regular, they have formulas.For arithmetic sequences, the common difference isd, and the first terma1is often referred to simply as"a". Since you get the next term by adding the common difference, the value ofa2is justa+d. The third term isa3= (a+d) +d=a+ 2d. The fourth term isa4= (a+ 2d) +d=a+ 3d. Following this pattern, then-th termanwill have the forman=a+ (n1)d.For geometric sequences, the common ratio isr, and the first terma1is often referred to simply as"a". Since you get the next term by multiplying by the common ratio, the value ofa2is justar. The third term isa3=r(ar) =ar2. The fourth term isa4=r(ar2) =ar3. Following this pattern, then-th termanwill have the forman=ar(n 1). Find the tenth term and then-th term of the following sequence:1/2, 1, 2, 4, 8,...The differences don't match:2 1 = 1, but4 2 = 2. So this isn't an arithmetic sequence. On the other hand, the ratios are the same:2 1 = 2, 4 2 = 2, 8 4 = 2. So this is a geometric sequence with common ratior= 2anda= 1/2. To find the tenth andn-th terms, I can just plug into the formulaan=ar(n 1):an= (1/2) 2n1a10= (1/2) 2101= (1/2) 29= (1/2)(512) = 256 Find then-th term and the first three terms of the arithmetic sequence havinga6= 5andd= 3/2.Then-th term of an arithmetic sequence is of the forman=a+ (n1)d. In this case, that formula gives mea6=a+ (6 1)(3/2) = 5. Solving this formula for the value of the first term of the sequence, I geta= 5/2. Then:a1= 5/2,a2= 5/2 + 3/2 = 1,a3= 1 + 3/2 = 1/2,andan= 5/2 + (n1)(3/2) Find then-th term and the first three terms of the arithmetic sequence havinga4= 93anda8= 65.Sincea4anda8are four places apart, then I know from the definition of an arithmetic sequence thata8=a4+ 4d. Using this, I can then solve for the common differenced:65 = 93 + 4d28 = 4d7 =dAlso, I know thata4=a+ (4 1)d, so, using the value I just found ford, I can find the value of the first terma:93 =a+ 3(7)93 + 21 =a114 =aOnce I have the value of the first term and the value of the common difference, I can plug-n-chug to find the values of the first three terms and the general form of then-th term:a1= 114,a2= 114 7 = 107,a3= 107 7 = 100an= 114 + (n1)(7) Find then-th and the26th terms of the geometric sequence witha5= 5/4anda12= 160.These two terms are12 5 = 7places apart, so, from the definition of a geometric sequence, I know thata12= (a5)(r7). I can use this to solve for the value of the common ratior:160 = (5/4)(r7)128 =r72 =rSincea5=ar4, then I can solve for the value of the first terma:5/4 =a(24) = 16a5/64 =aOnce I have the value of the first term and the value of the common ratio, I can plug each into the formulas, and find my answers:an= (5/64)2(n 1)a26= (5/64)(225) = 2 621 440Arithmetic Series(page 4 of 5)Sections:Terminology and notation,Basic examples,Arithmetic and geometric sequences, Arithmetic series,Finite and infinite geometric series

An arithmetic series is the sum of an arithmetic sequence. A geometric series is the sum of a geometric sequence. There are other types of series, but you're unlikely to work with them until you're in calculus. For now, you'll probably just work with these two.For reasons that will be explained in calculus, you can only take the partial sum of an arithmetic sequence. The "partial" sum is the sum of a limited (that is to say, finite) number of terms, like the first ten terms, or the fifth through the hundredth terms.The formula for the firstnterms of anarithmeticsequence, starting withn= 1,is:

The sum is, in effect,ntimes the "average" of the first and last terms. This sum of the firstnterms is called "then-th partial sum". (By the way: The above summation formulacan be provedusinginduction.) Find the35th partial sum ofan= (1/2)n+ 1The35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are:a1= (1/2)(1) + 1 = 3/2a2= (1/2)(2) + 1 = 2a3= (1/2)(3) + 1 = 5/2The terms have a common differenced= 1/2, so this is indeed an arithmetic sequence. The last term in the partial sum will bea35=a1+ (35 1)(d) = 3/2 + (34)(1/2) = 37/2. Then, plugging into the formula, the35th partial sum is:(n/2)(a1+an) = (35/2)(3/2 + 37/2) = (35/2)(40/2) =350 Find the value of the following summation:ADVERTISEMENT

From the formula ("2n 5") for then-thterm, I can see that each term will be two units larger than the previous term. (Plug in values fornif you're not sure about this.) So this is indeed an arithmetical sum. But this summation starts atn= 15, not atn= 1, and the summation formula applies to sums starting atn= 1. So how can I work with this summation?The quickest way to find the value of this sum is to find the14th and47th partial sums, and then subtract the14th from the47th. By doing this subtraction, I'll be left with the value of the sum of the15th through47th terms. The first term isa1= 2(1) 5 = 3. The other necessary terms area14= 2(14) 5 = 23anda47= 2(47) 5 = 89.

Subtracting, I get:

Then the solution is:Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

By the way, another notation for the summation of the first fourteen terms is "S14", so the subtraction could also be expressed as "S47 S14".Formatting note: Since this was just a summation, it's safe to assume that "2n 5" is the expression being summed. However (and especially if you're dealing with something more complex), sometimes grouping symbols may be necessary to make the meaning clear: Find the value ofnfor which the following equation is true:

I know that the first term isa1= 0.25(1) + 2 = 2.25. I can see from the formula that each term will be0.25units bigger than the previous term, so this is an arithmetical series. Then the summation formula for arithmetical series gives me:(n/2)(2.25 + [0.25n+ 2]) = 21n(2.25 + 0.25n+ 2) = 42n(0.25n+ 4.25) = 420.25n2+ 4.25n42 = 0n2+ 17n168 = 0(n+ 24)(n7) = 0Solving the quadratic, I get thatn= 24(which won't work in this context) orn= 7.n= 7Youcoulddo the above exercise by adding terms until you get to the required total of "21". But your instructor could easily give you a summation that requires, say, eighty-six terms before you get the right total. So make sure you can do the computations from the formula. Find the sum of1 + 5 + 9 + ... + 49 + 53.Checking the terms, I can see that this is indeed an arithmetic series:5 1 = 4, 9 5 = 4, 53 49 = 4. I've got the first and last terms, but how many terms are there in total?I have then-th term formula,"an=a1+ (n1)d", and I havea1= 1andd= 4. Plugging these into the formula, I can figure out how many terms there are:an=a1+ (n1)d53 = 1 + (n1)(4)53 = 1 + 4n453 = 4n356 = 4n14 =nSo there are14terms in this series. Now I have all the information I need:1 + 5 + 9 + ... + 49 + 53 = (14/2)(1 + 53) = (7)(54) =378Geometric Series(page 5 of 5)Sections:Terminology and notation,Basic examples,Arithmetic and geometric sequences,Arithmetic series, Finite and infinite geometric series

You can take the sum of a finite number of terms of a geometric sequence. And, for reasons you'll study in calculus, you can take the sum of aninfinitegeometric sequence, but onlyin the special circumstance that the common ratioris between1and1; that is, you have to have|r| < 1.For ageometric sequencewith first terma1=aand common ratior, the sum of the firstnterms is given by:

Note: Your book may have a slightly different form of the partial-sum formula above. For instance, the "a" may be multiplied through the numerator, the factors in the fraction might be reversed, or the summation may start ati= 0and have a power ofn+ 1on the numerator. All of these forms are equivalent, and the formulation abovemay be derivedfrompolynomial long division.In the special case that|r| < 1, the infinite sum exists and has the following value:Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

Evaluate the following:

The first few terms are6, 12, 24, so this is a geometric series with common ratior= 2. (I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of2.) The first term of the sequence isa= 6. Plugging into the summation formula, I get:

Sothe value of the summation is2 097 150 EvaluateS10for250, 100, 40, 16,....The notation "S10" means that I need to find the sum of the first ten terms. The first term isa= 250. Dividing pairs of terms, I get100 250 = 2/5, 40 100 = 2/5, etc, so the terms being added form a geometric sequence with common ratior= 2/5. When I plug in the values of the first term and the common ratio, the summation formula gives me:ADVERTISEMENT

Note: If you try to do the above computations in your calculator, it may very well return the decimal approximation of416.62297...instead of the fractional (and exact) answer. As you can see in the screen-capture to the right, entering the values in fractional form and using the "convert to fraction" command still results in just a decimal approximation to the answer. But (warning!) the decimal approximation will almost certain be regarded as a "wrong" answer! Take the time to find the fractional form!

FindanifS4= 26/27andr= 1/3.They've given me the sum of the first four terms,S4, and the value of the common ratior. Since there is a common ratio, I know this must be a geometric series. Plugging into the geometric-series-sum formula, I get:

Multiplying on both sides by27/40to solve for the first terma=a1, I get:

Then:

Show, by use of a geometric series, that0.3333...is equal to1/3.There's a trick to this. I first have to break the repeating decimal into separate terms:0.333... = 0.3 + 0.03 + 0.003 + 0.0003 + ...This shows the repeating pattern of the non-terminating (never-ending) decimal explicitly: For each term, I have a decimal point, followed by a steadily-increasing number of zeroes, and then ending with a "3". This can be written in fractional form, and then converted into geometric-series form:

Then0.333...is an infinite geometric series witha= 3/10andr= 1/10. Since|r| < 1, I can use the formula for summing infinite geometric series:

Using the summation formula to show that the geometric series "expansion" of0.333...has a value of one-third is the "showing" that the exercise asked for. You can use this method to convert any repeating decimal to its fractional form: By use of a geometric series, convert1.363636...to fractional form.First I'll break this into its constituent parts, so I can find the pattern:1.363636.. = 1 + 0.36 + 0.0036 + 0.000036 + ...There are two digits that repeat, so the fractions are a little bit different. But this is still a geometric series:

Then this is the leading "1", plus a geometric series havinga= 9/25andr= 1/100. Then the sum is:

Note: This technique can also be used to convert any repeating decimal into fractional form, and also can be used to prove that0.999... = 1.