Infinite Series

If {an}{a_n} is an infinite sequence, then

n=1an=a1+a2+a3+a4++an+\sum_{n=1}^{\infty} a_n = a_1+a_2+a_3+a_4+\ldots+a_n+\ldots

is an infinite series. If an infinite series has a finite sum, it is convergent. Otherwise, it is divergent.

For instance, in the last section, we saw the example from Zeno's Paradox, where we decided that the following must be true: n=112n=12+14+18+116+132+=1\sum_{n=1}^{\infty} \dfrac{1}{2^n} = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\ldots=1

We didn't have any justification for this, other than our intuition, but in this section, we'll see it proven.

It should be clear that in order for this to be possible, the terms in the series must be getting infinitesimally small, so that adding up an infinite number of terms won't blow up to an infinite answer. Is that enough, though? We'll see (specifically when we get to the harmonic series) that the answer is no. There are some series where the terms are getting infinitesimally small, but they aren't getting small fast enough, so the series diverges.

Convergence: Sequence of Partial Sums

Continue with the example of the series n=112n.\sum_{n=1}^{\infty} \dfrac{1}{2^n}. How can we add up an infinite number of terms? Well, let's start with the first term: s1=12.s_1=\dfrac{1}{2}. Then, add up the first two terms: s2=12+14=34.s_2=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}.

What if we add up the first three terms? Or the first 4? The first 20, or 100 terms? The point is this: if, as we add up more and more terms, the sum starts to approach a limit, we can be confident that the infinite series adds up to that number. If, on the other hand, it doesn't approach a limit, we can conclude that the series diverges.

Following this process uses what's called the sequence of partial sums, {sn}{s_n}:

s1=12s2=12+14=34s3=12+14+18=78s4=12+14+18+116=1516s5=12+14+18+116+132=3132\begin{aligned} s_1 &= \dfrac{1}{2}\ s_2 &= \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}\ s_3 &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} = \dfrac{7}{8}\ s_4 &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} = \dfrac{15}{16}\ s_5 &= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} = \dfrac{31}{32}\ &\vdots \end{aligned}

Let's look for a pattern in {sn}{s_n}: {sn}=12,34,78,1516,3132,{s_n}=\dfrac{1}{2},\dfrac{3}{4},\dfrac{7}{8},\dfrac{15}{16},\dfrac{31}{32},\ldots

Notice that the numerator is always 11 less than the denominator, and the denominators are the powers of 2.2. Therefore, the general formula for the SOPS is sn=2n12n.s_n=\dfrac{2^n−1}{2^n}.

Does this sequence of partial sums approach a limit? It does, and we can show this by splitting it into two fractions: limn2n12n=limn2n2n12n=limn112n=1\lim_{n \to \infty} \dfrac{2^n−1}{2^n} = \lim_{n \to \infty} \dfrac{2^n}{2^n}−\dfrac{1}{2^n} = \lim_{n \to \infty}1−\dfrac{1}{2^n}=1

Therefore, by using the SOPS, we've proven that n=112n=1\sum_{n=1}^{\infty} \dfrac{1}{2^n} = 1 just like we expected.


Given an infinite series a1+a2+a3+a4+a_1+a_2+a_3+a_4+\ldots the sequence of partial sums is given by {sn},{sn}, where sn=a1+a2++an.s_n=a_1+a_2+\ldots+a_n. If the sequence {sn}{s_n} is convergent and limnsn=s,\displaystyle\lim_{n \to \infty} s_n=s, then the series converges to ss as well: n=1an=s.\sum_{n=1}^{\infty} a_n=s.

Otherwise, the series is divergent.

We use the sequence of partial sums to define what it means for a series to converge (and it's a fairly intuitive definition, since we're essentially trying to add up an infinite list of numbers by adding up more and more each time), but we don't often use it to actually identify whether or not a series converges, because the SOPS doesn't typically fit such a nice pattern. However, I'll show one more example here where it can be used.

Determine whether the following series converges. n=1(1)n\sum_{n=1}^{\infty} (−1)^n


The series looks like n=1(1)n=1+11+11+1\sum_{n=1}^{\infty} (−1)^n=−1+1−1+1−1+1−\ldots

You should verify that the sequence of partial sums, therefore, looks like s1=1s2=0s3=1s4=0\begin{aligned} s_1 &= -1\ s_2 &= 0\ s_3 &= -1\ s_4 &= 0\ &\vdots \end{aligned}

Since this sequence clearly does not converge, but simply oscillates, we can conclude that n=1(1)n is divergent.\sum_{n=1}^{\infty}(−1)^n \textrm{ is divergent.}

The Harmonic Series

The harmonic series is a specific series: n=11n=1+12+13+14+15+16+17+18+19+\sum_{n=1}^{\infty} \dfrac{1}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\ldots

At first, after seeing the series at the beginning of the section, you may be tempted to conclude that this one converges as well. After all, the terms are getting smaller and smaller, infinitesimally small as the series goes on. However, it turns our that this series is divergent.

The Harmonic Series Diverges

Here's a "proof": start by writing out the first few terms of the series. n=11n=1+12+13+14+15+16+17+18+19+\sum_{n=1}^{\infty} \dfrac{1}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\ldots

Next, group terms together as follows: n=11n=1+12+13+14+15+16+17+18+19+\sum_{n=1}^{\infty} \dfrac{1}{n}=1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{3}+\dfrac{1}{4}}+\ans{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}}+\dfrac{1}{9}+\ldots

Replace the fractions in the boxes with smaller fractions that will make each box sum to 1/21/2 and collapse them: n=11n=1+12+13+14+15+16+17+18+19+>1+12+14+14+18+18+18+18+116+=1+12+12+12+\begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n} &= 1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{3}+\dfrac{1}{4}}+\ans{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}}+\dfrac{1}{9}+\ldots\ &> 1+\ans{\dfrac{1}{2}}+\ans{\dfrac{1}{4}+\dfrac{1}{4}}+\ans{\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}}+\dfrac{1}{16}+\ldots\ &= 1 + \ans{\dfrac{1}{2}} + \ans{\dfrac{1}{2}} + \ans{\dfrac{1}{2}} + \ldots \end{aligned}

It should be clear that this series diverges, because there will be an infinite number of 1/21/2s being added together. Since the harmonic series is larger than this divergent one, the harmonic series must diverge as well.

The harmonic series is a good example to remember, because there are cases where it can be used to check a result later.

Telescoping Series

Telescoping series aren't that common, but I'll show you an example here. In general, telescoping series have the form n=11n1n+C\sum_{n=1}^{\infty} \dfrac{1}{n}−\dfrac{1}{n+C} for some constant C.C. For example, consider the series n=11n1n+2=113+1214+1315+1416+1517+\sum_{n=1}^{\infty} \dfrac{1}{n}−\dfrac{1}{n+2}=1−\dfrac{1}{3}+\dfrac{1}{2}−\dfrac{1}{4}+\dfrac{1}{3}−\dfrac{1}{5}+\dfrac{1}{4}−\dfrac{1}{6}+\dfrac{1}{5}−\dfrac{1}{7}+\ldots

Notice what happens: the terms start to cancel each other, and the whole series collapses down (like a telescope; hence the name). 113+1214+1315+1416+1517+=1+12=32\begin{aligned} 1 &− \dfrac{1}{3}+\dfrac{1}{2}−\dfrac{1}{4}+\dfrac{1}{3}−\dfrac{1}{5}+\dfrac{1}{4}−\dfrac{1}{6}+\dfrac{1}{5}−\dfrac{1}{7}+\ldots\ &= 1+\dfrac{1}{2}\ &= \dfrac{3}{2} \end{aligned}

Therefore, this series converges: n=11n1n+2=32\sum_{n=1}^{\infty} \dfrac{1}{n} − \dfrac{1}{n+2} = \dfrac{3}{2}

Geometric Series


A geometric series is one in which each term is a consistent multiple of the previous term. For example, the series 4+43+49+427+=n=14(13)n4+\dfrac{4}{3}+\dfrac{4}{9}+\dfrac{4}{27}+\ldots = \sum_{n=1}^{\infty}4\left(\dfrac{1}{3}\right)^n is a geometric series, because each term is one third of the previous term.

In general, a geometric series has the form a+ar+ar2+ar3+ar4+=n=1arn1=n=0arn.a+ar+ar^2+ar^3+ar^4+\ldots = \sum_{n=1}^{\infty} ar^{n−1} = \sum_{n=0}^{\infty} ar^n.


The key to whether or not a geometric series converges is the value of the constant multiple rr: if r=1r=1 or r=1,r=−1, the series diverges, because it either adds the same term forever (tending toward infinity), or it oscillates between some value and zero. If r>1,|r|>1, the series also diverges, because the terms are getting bigger and bigger, so there's no hope of convergence in that case. The only case in which a geometric series converges is if r<1,|r|<1, in which case the series converges to a1r.\dfrac{a}{1−r}. (proof)

Summary: If r1, then the series is divergent. If r<1, then the series is convergent, andn=1arn1=a1r\ans{\begin{aligned} \textrm{Summary:} &\textrm{ If } |r| \geq 1 \textrm{, then the series is divergent.}\ &\textrm{ If } |r| < 1 \textrm{, then the series is convergent, and}\ &\sum_{n=1}^{\infty} ar^{n-1} = \dfrac{a}{1-r} \end{aligned}}

Determine whether the following series converges. n=14(13)n1\sum_{n=1}^{\infty} 4\left(\dfrac{1}{3}\right)^{n−1}


Here, a=4a=4 and r=1/3.r=1/3. Since r<1,|r|<1, this series is convergent: n=14(13)n1=411/3=6\sum_{n=1}^{\infty} 4\left(\dfrac{1}{3}\right)^{n−1}=\dfrac{4}{1−1/3}=\ans{6}

Determine whether the following series converges. 5103+2094027+5−\dfrac{10}{3}+\dfrac{20}{9}−\dfrac{40}{27}+\ldots


The first term in the geometric series is a,a, so in this example, a=5.a=5. To get from one term to the next, multiply by 2/3,−2/3, so r=2/3.r=−2/3. Since r<1,|r|<1, this series is convergent: n=15(23)n1=51+2/3=3\sum_{n=1}^{\infty} 5\left(−\dfrac{2}{3}\right)^{n−1}=\dfrac{5}{1+2/3}=\ans{3}

Determine whether the following series converges. n=122n 31n\sum_{n=1}^{\infty} 2^{2n}\ 3^{1−n}


This one is also a geometric series, although it's not written in standard form. To get it in standard form, we could manipulate it algebraically, but it's easier to simply list the first few terms, from which we can extract aa and rr; remember, the first term is aa and the multiple is r.r. n=122n 31n=4+163+649+\sum_{n=1}^{\infty} 2^{2n}\ 3^{1−n}=4+\dfrac{16}{3}+\dfrac{64}{9}+\ldots

Therefore, a=4a=4 and r=4/3.r=4/3. Since r>1,|r|>1, this series is divergent: n=122n 31n is divergent.\ans{\sum_{n=1}^{\infty} 2^{2n}\ 3^{1−n} \textrm{ is divergent.}}

  1. Determine whether the following series converges. If it does, find its sum. 3+2+43+89+3+2+\dfrac{4}{3}+\dfrac{8}{9}+\ldots

  2. Converges to 9

  3. Determine whether the following series converges. If it does, find its sum. 1814+121+\dfrac{1}{8}−\dfrac{1}{4}+\dfrac{1}{2}−1+\ldots

  4. Diverges

  5. Determine whether the following series converges. If it does, find its sum. n=11n1n+1\sum_{n=1}^{\infty} \dfrac{1}{n}−\dfrac{1}{n+1}

  6. Converges to 1

  7. Determine whether the following series converges. If it does, find its sum. n=11+2n3n\sum_{n=1}^{\infty} \dfrac{1+2^n}{3^n}

  8. Converges to 92\dfrac{9}{2}

  9. Determine whether the following series converges. If it does, find its sum. n=132n422n\sum_{n=1}^{\infty} 3^{2n} 4^{2−2n}

  10. Converges to 1447\dfrac{144}{7}