The Comparison Test

We have already used the Monotone Convergence Theorem in studying simple series. In fact it is a lot more useful. When we know the behaviour of some simple series, we can deduce many more results by comparison as follows.

Theorem 6.12   (The Comparison Test)Assume that 0$\le$a_n$\le$b_n for all n, and suppose that $\sum$b_n is convergent. Then $\sum$a_n is convergent.

Proof. Define

S_n = a₁ + a₂ + ^... + a_n,

T_n = b₁ + b₂ + ^... b_n.

Then by hypothesis, 0$\le$S_n$\le$T_n. Since {T_n} is a convergent sequence, it is a bounded sequence by Prop 2.28. In particular, it is bounded above, so there is some K such that T_n$\le$K for all n. Thus S_n$\le$K for all n, so {S_n} is a sequence that is bounded above; since a_n$\ge$ 0, S_n+1 = S_n + a_n$\ge$S_n and {S_n} is an increasing sequence. Thus by the Monotone Convergence Theorem, it is a convergent sequence $\qedsymbol$

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Example 6.13   Let a_n = $${\frac{{2n}}{{3n^3-1}}}$$ and let b_n = $${\frac{{1}}{{n^2}}}$$. Then $\sum$a_n is convergent

For n$\ge$1, n³$\ge$1, so 3n³ - 1$\ge$2n³. Thus

a_n = $${\frac{{2n}}{{3n^3-1}}}$$$$\le$$$${\frac{{2n}}{{2n^3}}}$$ = b_n.

Since we know that $\sum$b_n is convergent, so is $\sum$a_n.

Remark 6.14   The conclusions of Theorem 6.12 remain true even if we only have a_n$\le$b_n eventually; for if it holds for n$\ge$N, we replace the inequality by

S_n$$\le$$T_n + a₁ + a₂ + ^... + a_N

and this then holds for all n.

Example 6.15   Let a_n = $${\frac{{\log n}}{{n}}}$$, and compare with b_n = $${\frac{{1}}{{n}}}$$.

Solution. Note that if n$\ge$3, then log n > 1. We can thus use the ``eventually'' form of the comparison test; we have

a_n = $${\frac{{\log n}}{{n}}}$$ > $${\frac{{1}}{{n}}}$$ = b_n,

We deduce divergence, for if $\sum$a_n were convergent, it follows that that $\sum$b_n was convergent, which it isn't!

Corollary 6.16 (Limiting form of the Comparison Test)   Suppose that a_n > 0 and b_n > 0, and that there is some constant k such that $$\lim_{{n\to\infty}}^{}$$$${\frac{{a_n}}{{b_n}}}$$ = k > 0. Then $\sum$a_n is convergent iff $\sum$b_n is convergent.

Proof. Assume first that $\sum$b_n is convergent. Since a_n/b_n$\to$k as n$\to$$\infty$, eventually (take $\epsilon$ = k > 0), we have a_n$\le$2kb_n, which is convergent by 6.11. Hence $\sum$a_n is convergent by 6.12. To get the converse, note that b_n/a_n$\to$1/k as n$\to$$\infty$, so we can use the same argument with a_n and b_n interchanged. $\qedsymbol$

Example 6.17   Let a_n = $${\frac{{n}}{{n^2+1}}}$$ and let b_n = $${\frac{{1}}{{n}}}$$. Then $\sum$a_n is divergent by the limiting form of the comparison test.

Solution. Note that the terms are all positive, so we try to apply the limiting form of the comparison test directly.

$${\frac{{a_n}}{{b_n}}}$$ = $${\frac{{n}}{{n^2+1}}}$$.$${\frac{{n}}{{1}}}$$ = $${\frac{{n^2}}{{n^2 + 1}}}$$$$\to$$1    as n$\to$$\infty$.    

Since the limit is non-zero, the use of the limiting form of the comparison test is valid, and we see that $\sum$a_n is divergent.

Note also we need our work on sequences in Section 2 to evaluate the required limit.

This is all very well, but as with the ``new sequences from old'' programme, we need a few reference sequences before we can get further. One set is the geometric series, which we have already met.

Exercise 6.18   Let a_n = $${\frac{{2^n+7}}{{3^n-1}}}$$ and let b_n = $$\left(\vphantom{\frac{2}{3}}\right.$$$${\frac{{2}}{{3}}}$$$$\left.\vphantom{\frac{2}{3}}\right)^{n}_{}$$. Use the limiting form of the comparison test to show that $\sum$a_n is convergent.

We also know about $\sum$1/n^$\alpha$, at least when $\alpha$$\ge$2 when it converges, by comparison with $\sum$1/n², and when $\alpha$$\le$1 when it diverges, by comparison with $\sum$1/n.

Proposition 6.19   The sum $$\sum$$$${\frac{{1}}{{n^{\alpha}}}}$$ is convergent when $\alpha$ > 1.

Solution. Assume $\alpha$ > 1; we estimate the partial sums. Since 1/n^$\alpha$ > 0, clearly {S_n} is an increasing sequence. Let S_n = 1 + $${\frac{{1}}{{2^\alpha}}}$$ + ^... + $${\frac{{1}}{{n^\alpha}}}$$, and consider the graph of y = 1/x^$\alpha$, noting that y is a decreasing function of x (which is where we use that fact that $\alpha$ > 1). From a diagram which is essentially the same as that of Fig 6.1,

$$\int_{1}^{n} {\frac{{d\,x}}{{x^\alpha}}}$$     shaded area    

$${\frac{{1}}{{2^\alpha}}} {\frac{{1}}{{n^\alpha}}} \left[\vphantom{\frac{1}{1-\alpha}\frac{1}{x^{\alpha-1}}}\right. {\frac{{1}}{{1-\alpha}}} {\frac{{1}}{{x^{\alpha-1}}}} \left.\vphantom{\frac{1}{1-\alpha}\frac{1}{x^{\alpha-1}}}\right]^{n}_{1}$$

$${\frac{{1}}{{\alpha-1}}} \left(\vphantom{1-\frac{1}{n^{\alpha-1}}}\right. {\frac{{1}}{{n^{\alpha-1}}}} \left.\vphantom{1-\frac{1}{n^{\alpha-1}}}\right)$$ S_n - 1

$${\frac{{1}}{{\alpha-1}}}$$ S_n

Thus the sequence of partial sums is bounded above, and the series converges.

Exercise 6.20   Let a_n = $${\frac{{n}}{{\sqrt{n^5 + n + 1}}}}$$ and let b_n = $${\frac{{1}}{{n^{3/2}}}}$$. Use the limiting form of the comparison test to show that $\sum$a_n is convergent.

We can consider the method of comparing with integrals as an ``integral test'' for the convergence of a series; rather than state it formally, note the method we have used.

Theorem 6.21 (The Ratio Test)   Let $\sum$a_n be a series, and assume that lim$${\frac{{\vert a_{n+1}\vert}}{{\vert a_n\vert}}}$$$$\to$$r as n$\to$$\infty$. Then if r < 1, the series is convergent, if r > 1, the series is divergent, while if r = 1, the test gives no information.

Proof. A proof follows by comparing with the corresponding geometric series with ratio r. Details will be given in full in the third year course. $\qedsymbol$

Example 6.22   Let a_n = $${\frac{{2^n(n!)^2}}{{(2n)!}}}$$. Then $\sum$a_n is convergent.

Solution. We look the ratio of adjacent terms in the series (of positive terms).

$${\frac{{\hbox{$(n+1)^{\rm st}$\ term}}}{{\hbox{$n^{\rm th}$\ term}}}} {\frac{{a_{n+1}}}{{a_n}}} {\frac{{2^{n+1}(n+1)!(n+1)!}}{{(2n+2)!}}} {\frac{{(2n)!}}{{2^n n!n!}}}$$

$${\frac{{2(n+1)^2}}{{(2n+2)(2n+1)}}} {\frac{{n+1}}{{2n+1}}} \to {\frac{{1}}{{2}}} \to \infty$$

Since the ratio of adjacent terms in the series tends to a limit which is < 1, the series converges by the ratio test.