Convergent Series

Definition 6.3   Let {a_n} be a sequence, and let

S_n = a₁ + a₂ + ^... + a_n,    the n^th partial sum.    

If $$\lim_{{n\to\infty}}^{}$$S_n exists, we say that $$\sum$$a_n is a convergent series, and write $$\lim_{{n\to\infty}}^{}$$S_n = $$\sum$$a_n.

Thus a series is convergent if and only if it's sequence of partial sums is convergent. The limit of the sequence of partial sums is the sum of the series. A series which is not convergent, is a divergent series.

Example 6.4   The series $$\sum$$rⁿ is convergent with sum 1/(1 - r), provided that | r| < 1. For other values of r, the series is divergent; in particular, the series $$\sum$$(- 1)ⁿ is divergent.

Solution. We noted above that when | r| < 1, S_n$\to$a/(1 - r) as n$\to$$\infty$; note particular cases;

$$\sum_{{n=1}}^{\infty}$$$${\frac{{1}}{{2^n}}}$$ = 1    or equivalently,    $${\frac{{1}}{{2}}}$$ + $${\frac{{1}}{{4}}}$$ + $${\frac{{1}}{{8}}}$$ + ^... = 1.

Example 6.5   The sum $$\sum$$$${\frac{{1}}{{n(n+1)}}}$$ is convergent with sum 1.

Solution. We can compute the partial sums explicitly:

S_n = $$\sum_{{k=1}}^{n}$$$${\frac{{1}}{{k(k+1)}}}$$ = $$\sum_{{k=1}}^{n}$$$$\left(\vphantom{\frac{1}{k} - \frac{1}{k+1}}\right.$$$${\frac{{1}}{{k}}}$$ - $${\frac{{1}}{{k+1}}}$$$$\left.\vphantom{\frac{1}{k} - \frac{1}{k+1}}\right)$$ = 1 - $${\frac{{1}}{{n+1}}}$$$$\to$$1    as    n$$\to$$$$\infty$$.

Example 6.6   The sum $$\sum$$$${\frac{{1}}{{n}}}$$ is divergent.

Solution. We estimate the partial sums:

$${\frac{{1}}{{1}}} {\frac{{1}}{{2}}} \left(\vphantom{\frac{1}{3}+ \frac{1}{4}}\right. {\frac{{1}}{{3}}} {\frac{{1}}{{4}}} \left.\vphantom{\frac{1}{3}+ \frac{1}{4}}\right) \left(\vphantom{\frac{1}{5}+ \cdots + \frac{1}{8}}\right. {\frac{{1}}{{5}}} {\frac{{1}}{{8}}} \left.\vphantom{\frac{1}{5}+ \cdots + \frac{1}{8}}\right) \left(\vphantom{\frac{1}{9} +\cdots+ \frac{1}{15}}\right. {\frac{{1}}{{9}}} {\frac{{1}}{{15}}} \left.\vphantom{\frac{1}{9} +\cdots+ \frac{1}{15}}\right) {\frac{{1}}{{n}}}$$ S_n =

$${\frac{{1}}{{2}}} {\frac{{2}}{{4}}} {\frac{{4}}{{8}}} {\frac{{1}}{{2}}} \ge$$ >

$${\frac{{1}}{{2}}} {\frac{{2}}{{4}}} {\frac{{4}}{{8}}} {\frac{{8}}{{16}}} \ge$$ >

$$\to \infty \to \infty$$

Example 6.7   The sum $$\sum$$$${\frac{{1}}{{n^2}}}$$ is convergent. [Actually the sum is $\pi^{2}_{}$/6, but this is much harder.]

Figure 6.1: Comparing the area under the curve y = 1/x² with the area of the rectangles below the curve

Solution. We estimate the partial sums. Since 1/n² > 0, clearly {S_n} is an increasing sequence. We show it is bounded above, whence by the Monotone Convergence Theorem (3.9), it is convergent. From the diagram,

$${\frac{{1}}{{2^2}}} {\frac{{1}}{{3^2}}} {\frac{{1}}{{n^2}}} \int_{1}^{n} {\frac{{d\,x}}{{x^2}}}$$ <

$$\left[\vphantom{-\frac{1}{x}}\right. {\frac{{1}}{{x}}} \left.\vphantom{-\frac{1}{x}}\right]_{1}^{n} \le {\frac{{1}}{{n}}}$$ S_n <

Thus S_n < 2 for all n, the sequence of partial sums is bounded above, and the series is convergent.

Proposition 6.8   Let $\sum$a_n be convergent. Then a_n$\to$ 0 as n$\to$$\infty$.

Proof. Write l = $\lim_{{n\to \infty}}^{}$S_n, and recall from our work on limits of sequences that S_n-1$\to$l as n$\to$$\infty$. Then

a_n = (a₁ + a₂ +...a_n) - (a₁ + a₂ +...a_n-1) = S_n - S_n-1$$\to$$l - l    as    n$$\to$$$$\infty$$.

$\qedsymbol$

Remark 6.9   This gives a necessarycondition for the convergence of a series; it is not sufficient. For example we have seen that $\sum$1/n is divergent, even though 1/n$\to$ 0 as n$\to$$\infty$.

Example 6.10   The sum $$\sum$$$${\frac{{1}}{{n}}}$$ is divergent (Graphical method).

Solution. We estimate the partial sums. Since 1/n > 0, clearly {S_n} is an increasing sequence. We show it is not bounded above, whence by the note after 3.9, the sequence of partial sums $\to$$\infty$ as n$\to$$\infty$.

Figure 6.2: Comparing the area under the curve y = 1/x with the area of the rectangles above the curve

From the diagram,

1 + $${\frac{{1}}{{2}}}$$ + ^... + $${\frac{{1}}{{n}}}$$ > $$\int_{1}^{n}$$$${\frac{{d\,x}}{{x}}}$$ > $${\frac{{1}}{{2}}}$$ + ^... + $${\frac{{1}}{{n}}}$$.

Writing S_n = 1 + $${\frac{{1}}{{2}}}$$ + ^... + $${\frac{{1}}{{n}}}$$, we have S_n > log n > S_n - 1, or equivalently 1 + log n > S_n > log n for all n. Thus S_n$\to$$\infty$ and the series is divergent. [There is a much better estimate; the difference S_n - log n$\to$$\gamma$ as n$\to$$\infty$, where $\gamma$ is Euler's constant.]

Proposition 6.11   Let $\sum$a_n and $\sum$b_n be convergent. Then $\sum$(a_n + b_n) and $\sum$c.a_n are convergent.

Proof. This can be checked easily directly from the definition; it is in effect the same proof that the sum of two convergent sequences is convergent etc. $\qedsymbol$