• Infinite series may begin with any index. For example, \[ \sum_{n=3}^\infty \frac1n = \frac13+\frac14+\frac15+\cdots \] When it is not necessary to specify the starting point, we write simply \({\sum a_n}\).
• Any letter may be used for the index. Thus, we may write \(a_m\), \(a_k\), \(a_i\) etc.

DEFINITION Convergence of an Infinite Series

An infinite series \(\sum\limits_{n=k}^{\infty} a_n\) converges to the sum \(S\) if its partial sums converge to \(S\): \[\lim\limits_{N\to\infty}S_N =S\]

In this case, we write \(S = \sum\limits_{n=k}^\infty a_n\).

• If the limit does not exist, we say that the infinite series diverges.
• If the limit is infinite, we say that the infinite series diverges to infinity.

EXAMPLE 1 Telescoping Series

Investigate numerically: \[ S=\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac1{1(2)} + \frac1{2(3)} + \frac1{3(4)}+\frac1{4(5)} + \cdots \]

Then compute the sum \(S\) using the identity: \[ \frac1{n(n+1)} = \frac1n-\frac1{n+1} \]

Solution The values of the partial sums listed in Table 10.2 suggest convergence to \(S = 1\) To prove this, we observe that because of the identity, each partial sum collapses down to just two terms: \[ \begin{eqnarray*} S_1 &=& \frac1{1(2)} = \frac11 - \frac12\\ S_2&=& \frac1{1(2)} + \frac1{2(3)} = \left(\frac11 - \cancel{\frac12}\right) +\left( \cancel{\frac12} - \frac13\right) = 1-\frac13\\ S_3 &=& \frac1{1(2)} + \frac1{2(3)} + \frac1{3(4)} = \left( \frac11 - \cancel{\frac12}\right) +\left( \cancel{\frac12} - \cancel{\frac13}\right) + \left(\cancel{\frac13}-\frac14\right) = 1 - \frac14 \end{eqnarray*} \]

In general, \[ \begin{equation*} S_N =\left( \frac11 - \cancel{\frac12}\right) +\left( \cancel{\frac12} - \cancel{\frac13}\right) +\cdots + \left(\cancel{\frac1N}-\frac1{N+1}\right) = 1 - \frac1{N+1} \tag {2} \end{equation*} \]

The sum \(S\) is the limit of the partial sums: \[ S = \lim_{N\to\infty}S_N = \lim_{N\to\infty} \left(1 - \frac1{N+1}\right) = 1 \]

EXAMPLE 2 Sequences versus Series

Discuss the difference between \(\{a_n\}\) and \(\sum\limits_{n=1}^{\infty}a_n\), where \(a_n = \frac{1}{n(n+1)}\).

Solution The sequence is the list of numbers \({\frac1{1(2)}, \frac1{2(3)}, \frac1{3(4)},\ldots.}\) This sequence converges to zero: \[ \lim_{n\to \infty} a_n = \lim_{n\to \infty} \frac{1}{n(n+1)} = 0 \]

The infinite series is the sum of the numbers \(a_n\) defined formally as the limit of the partial sums. This sum is not zero. In fact, the sum is equal to 1 by Example 1: \[ \sum\limits_{n=1}^{\infty}a_n = \sum_{n=1}^\infty\,\frac1{n(n+1)} = \frac1{1(2)}+\frac1{2(3)}+\frac1{3(4)}+\cdots = 1 \]

THEOREM 1 Linearity of Infinite Series

If, \(\sum a_n\) and \(\sum b_n\) converge, then \(\sum (a_n \pm b_n)\) and \(\sum ca_n\) also converge (\(c\) any constant), and \[\begin{array}{*{20}{c}} \begin{align*} {\sum {a_n} + \sum {b_n}}{=}&{\sum ({a_n} + {b_n})} \\ {\sum {a_n} - \sum {b_n}}{=}&{\sum ({a_n} - {b_n})} \\ {\sum c{a_n}}{=}&{c\sum {a_n}\;\;\;(c~\textrm{any onstant})} \\ \end{align*}\end{array}\]

Proof

These rules follow from the corresponding linearity rules for limits. For example, \[\begin{array}{*{20}{c}} {\mathop \sum \limits_{n = 1}^\infty ({a_n} + {b_n})} & = & {\mathop {\lim }\limits_{N \to \infty } \mathop \sum \limits_{n = 1}^\infty ({a_n} + {b_n}) = \mathop {\lim }\limits_{N \to \infty } \left( {\mathop \sum \limits_{n = 1}^N {a_n} + \mathop \sum \limits_{n = 1}^N {b_n}} \right)} \\ {} & = & {\mathop {\lim }\limits_{N \to \infty } \mathop \sum \limits_{n = 1}^\infty {a_n} + \mathop {\lim }\limits_{N \to \infty } \mathop \sum \limits_{n = 1}^\infty {b_n} = \mathop \sum \limits_{n = 1}^\infty {a_n} + \mathop \sum \limits_{n = 1}^\infty {b_n}} \\ \end{array}\]

Geometric series are important because they

• arise often in applications.
• can be evaluated explicitly.
• are used to study other, nongeometric series (by comparison).

THEOREM 2 Sum of a Geometric Series

Let \(c\neq 0\) If \(|r|<1\) then \[ \begin{eqnarray*} \boxed{\bbox[#FAF8ED,5pt]{\sum\limits_{n=0}^\infty cr^n = c+cr+cr^2+cr^3+\cdots = \frac{c}{1-r} }}\tag{4}\\ \boxed{\bbox[#FAF8ED,5pt]{\sum\limits_{n=M}^\infty cr^n = cr^M+cr^{M+1}+cr^{M+2}+cr^{M+3} +\cdots = \frac{cr^M}{1-r} }}\tag{5} \end{eqnarray*} \] If \(|r|\ge 1\) then the geometric series diverges.

Proof

If \(r=1\) then the series certainly diverges because the partial sums \(S_N=N\!c\) grow arbitrarily large. If \(r\neq 1\) then Eq. (3) yields \[ S = \lim_{N\to\infty} S_N = \lim_{N\to\infty} \frac{c(1 - r^{N+1})}{1 - r} = \frac{c}{1 - r} - \frac{c}{1 - r} \lim_{N\to\infty} r^{N+1} \]

If \(|r|<1\) then, \(\lim\limits_{N\to\infty}r^{N+1}=0\) and we obtain Eq (4). If \(|r|\geq 1\) and \(r\neq1\) then \(\lim\limits_{N\to\infty}r^{N+1}\) does not exist and the geometric series diverges. Finally, if the series starts with \(cr^M\) rather than \(cr^0\) then \[ \begin{equation*} S = cr^M + cr^{M + 1} + cr^{M + 2} + cr^{M + 3} + \cdots = r^M\smash{\sum_{n = 0}^\infty cr^n = \frac{cr^M}{1-r}} \end{equation*} \]

EXAMPLE 3

Evaluate \(\sum\limits_{n=0}^{\infty} 5^{-n}\).

Solution This is a geometric series with \(r=5^{-1}\) By Eq.(4), \[ \sum_{n=0}^{\infty} 5^{-n} = 1 + \frac15 + \frac1{5^2}+ \frac1{5^3} + \cdots = \frac1{1-5^{-1}} = \frac54 \]

EXAMPLE 4

Evaluate \(\sum\limits_{n=3}^{\infty} 7\left(-\frac{3}{4}\right)^n= 7\left(-\dfrac{3}{4}\right)^3 + 7\left(-\dfrac{3}{4}\right)^4 + 7\left(-\dfrac{3}{4}\right)^5 + \cdots\).

Solution This is a geometric series with \(r=-\tfrac34\) and \(c=7\) starting at \(n=3\) By Eq. (5) , \[ \sum_{n=3}^{\infty}7\left(-\dfrac{3}{4}\right)^n = \frac{7\bigl(-\frac{3}{4}\bigr)^3}{1 - \bigl(-\frac34\bigr)} = -\frac{27}{16} \]

EXAMPLE 5

Evaluate \(\displaystyle S = \sum_{n = 0}^{\infty} \frac{2 + 3^n}{5^{n}}.\)

Solution Write \(S\) as a sum of two geometric series. This is valid by Theorem 1 because both geometric series converge: \[ \begin{align*} \sum_{n = 0}^{\infty} \frac{2 + 3^n}{5^n} &= \sum_{n = 0}^{\infty} \frac{2}{5^n} + \sum_{n = 0}^{\infty} \frac{3^n}{5^n} = \overbrace{2\sum_{n = 0}^{\infty} \frac{1}{5^n} + \sum_{n = 0}^{\infty} \left(\frac{3}{5}\right)^n}^{\textrm{Both geometric series converge}}\\ &= 2\cdot \frac{1}{1 - \frac{1}{5}} + \frac{1}{1 - \frac{3}{5}} = 5 \end{align*} \]

GRAPHICAL INSIGHT

Sometimes, the following incorrect argument is given for summing a geometric series: \[ \begin{eqnarray*} S &=& \phantom{1+{}} \frac12+\frac14+\frac18+\cdots\\ 2S &=& 1+\frac12+\frac14+\frac18+\cdots = 1+S \end{eqnarray*} \]

Thus, \(2S=1+S\) or \(S=1\) The answer is correct, so why is the argument wrong? It is wrong because we do not know in advance that the series converges. Observe what happens when this argument is applied to a divergent series: \[ \begin{eqnarray*} S &=& 1+2+4+8+16+\cdots \\ 2S &=& \phantom{1+{}} 2+4+8+16+\cdots = S-1 \end{eqnarray*} \]

This would yield \(2S=S-1\) or \(S=-1\) which is absurd because \(S\) diverges. We avoid such erroneous conclusions by carefully defining the sum of an infinite series as the limit of partial sums.

THEOREM 3 Divergence Test

If the \(n\) th term \(a_n\) does not converge to zero, then the series \(\sum\limits_{n=1}^\infty a_n\) diverges.

The Divergence Test (also called the nth-Term Test) is often stated as follows: \[ If \sum_{n=1}^\infty a_n~converges,~then \lim_{n\to\infty}a_n =0. \]

In practice, we use it to prove that a given series diverges.

Proof

First, note that \(a_n = S_n - S_{n-1}\) because \[ S_{n} = \big(a_1+a_2+\cdots +a_{n-1}\big)+a_n = S_{n-1} + a_n \]

If \(\sum\limits_{n=1}^\infty a_n\) converges with sum \(S\) then \[ \lim_{n\to\infty}a_n = \lim_{n\to\infty}(S_n-S_{n-1}) = \lim_{n\to\infty}S_n - \lim_{n\to\infty}S_{n-1}=S-S=0 \]

Therefore, if \(a_n\) does not converge to zero, \(\sum\limits^\infty_{n=1} a_n\) cannot converge.

EXAMPLE 6

Prove the divergence of \(S = \sum\limits_{n=1}^\infty \frac{n}{4n+1}\).

Solution We have \[ \lim_{n\to\infty}a_n = \lim_{n\to\infty} \frac{n}{4n+1} = \lim_{n\to\infty} \frac{1}{4+1/n} =\frac14 \]

The nth term \(a_n\) does not converge to zero, so the series diverges by Theorem 3.

EXAMPLE 7

Determine the convergence or divergence of \[ S = \sum_{n=1}^\infty (-1)^{n-1} \frac{n}{n+1} = \frac12 - \frac23 + \frac34 - \frac45 + \,\cdots \]

Solution The general term \(a_n = (-1)^{n-1} \frac{n}{n+1}\) does not approach a limit. Indeed, \(\frac{n}{n+1}\) tends to \(1\) so the odd terms \(a_{2n+1}\) tend to \(1\) and the even terms \(a_{2n}\) tend to \(-1\) Because \(\lim\limits_{n\to\infty}a_n\) does not exist, the series \(S\) diverges by Theorem 3.

EXAMPLE 8 Sequence Tends to Zero, yet the Series Diverges

Prove the divergence of \[ \sum\limits_{n=1}^\infty \frac1{\sqrt{n}} = \frac1{\sqrt 1} +\frac1{\sqrt 2}+\frac1{\sqrt 3}+\cdots \]

Solution The general term \({1}/{\sqrt{n}}\) tends to zero. However, because each term in the partial sum \(S_N\) is greater than or equal to \(1/\sqrt{N}\) we have \[ \begin{eqnarray*} S_N &=& \overbrace{\frac1{\sqrt 1} +\frac1{\sqrt 2}+ \cdots + \frac1{\sqrt N}}^{\textrm{\(N\) terms}}\\ &\ge& \frac1{\sqrt N} +\frac1{\sqrt N} + \cdots + \frac1{\sqrt N} \\&=& N\left(\frac{1}{\sqrt N}\right) = \sqrt{N} \end{eqnarray*} \]

This shows that \(S_N\ge \sqrt N\) But \(\sqrt{N}\) increases without bound (Figure 10.14). Therefore \(S_N\) also increases without bound. This proves that the series diverges.

HISTORICAL PERSPECTIVE

Geometric series were used as early as the third century BCE by Archimedes in a brilliant argument for determining the area \(S\) of a “parabolic segment” (shaded region in Figure 10.15). Given two points \(A\) and \(C\) on a parabola, there is a point \(B\) between \(A\) and \(C\) where the tangent line is parallel to \(\overline{AC}\) (apparently, Archimedes knew the Mean Value Theorem more than \(2000\) years before the invention of calculus). Let \(T\) be the area of triangle \(\triangle ABC\) Archimedes proved that if \(D\) is chosen in a similar fashion relative to \(\overline{AB}\) and \(E\) is chosen relative to \(\overline{BC}\) then \[ \begin{eqnarray*} \frac14 T &=& \textrm{Area}(\triangle ADB) + \textrm{Area}(\triangle BEC)\tag {6} \end{eqnarray*} \]

This construction of triangles can be continued. The next step would be to construct the four triangles on the segments \(\overline{AD}\) \(\overline{DB}\) \(\overline{BE}\) \(\overline{EC}\) of total area \(\frac14^2T\) Then construct eight triangles of total area \(\frac14^3T\) etc. In this way, we obtain infinitely many triangles that completely fill up the parabolic segment. By the formula for the sum of a geometric series, \[ S = T + \frac14T+\frac{1}{16}T + \cdots = T\sum_{n=0}^\infty \frac1{4^n} = \frac43 T \]

For this and many other achievements, Archi-medes is ranked together with Newton and Gauss as one of the greatest scientists of all time.

The modern study of infinite series began in the seventeenth century with Newton, Leibniz, and their contemporaries. The divergence of \(\sum\limits_{n=1}^\infty 1/n\) (called the harmonic series) was known to the medieval scholar Nicole d’Oresme (1323–1382), but his proof was lost for centuries, and the result was rediscovered on more than one occasion. It was also known that the sum of the reciprocal squares \( \sum\limits_{n=1}^\infty 1/{n^2}\) converges, and in the 1640s, the Italian Pietro Mengoli put forward the challenge of finding its sum. Despite the efforts of the best mathematicians of the day, including Leibniz and the Bernoulli brothers Jakob and Johann, the problem resisted solution for nearly a century. In 1735, the great master Leonhard Euler (at the time, 28 years old) astonished his contemporaries by proving that \[ \begin{equation*} \boxed{\bbox[#FAF8ED,5pt]{\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{5^2}}} + \cdots = \frac{{{\pi ^2}}}{6}}}\end{equation*}\]

This formula, surprising in itself, plays a role in a variety of mathematical fields. A theorem from number theory states that two whole numbers, chosen randomly, have no common factor with probability \(6/\pi^2\approx 0.6\) (the reciprocal of Euler’s result). On the other hand, Euler’s result and its generalizations appear in the field of statistical mechanics.

Archimedes (287 BCE–212 BCE), who discovered the law of the lever, said “Give me a place to stand on, and I can move the earth” (quoted by Pappus of Alexandria c. AD 340).