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Calculus
Concept Version 7
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Convergence of Series with Positive Terms

For a sequence {an}\{a_n\}{a​n​​}, where ana_na​n​​ is a non-negative real number for every nnn, the sum ∑n=0∞an\sum_{n=0}^{\infty}a_n∑​n=0​∞​​a​n​​ can either converge or diverge to ∞\infty∞.

Learning Objective

  • Identify convergence conditions for a sequence with positive terms


Key Points

    • Because the partial sum SkS_kS​k​​ of a series with non-negative terms can only increase as kkk becomes larger, the limit of the partial sum can either converge or diverge to ∞\infty∞.
    • A geometric sum 1+12+14+18+⋯+12n+⋯1 + \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots+ \frac{1}{2^n}+\cdots1+​2​​1​​+​4​​1​​+​8​​1​​+⋯+​2​n​​​​1​​+⋯ converges to 222, which can be understood visually.
    • The series ∑n≥11n2\sum_{n \ge 1} \frac{1}{n^2}∑​n≥1​​​n​2​​​​1​​ is convergent. This can be seen by comparing individual terms of the series with a sequence (bn=1n−1−1n)\left( b_n = \frac{1}{n-1} - \frac{1}{n} \right)(b​n​​=​n−1​​1​​−​n​​1​​), which is know to converge.

Terms

  • converge

    of a sequence, to have a limit

  • convergence test

    methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence, or divergence of an infinite series


Full Text

For a sequence {an}\{a_n\}{a​n​​}, where ana_na​n​​ is a non-negative real number for every nnn, the sequence of partial sums

Sk=∑n=0kan=a0+a1+⋯+akS_k = \sum_{n=0}^{k}a_n = a_0 + a_1 + \cdots + a_kS​k​​=∑​n=0​k​​a​n​​=a​0​​+a​1​​+⋯+a​k​​ 

is non-decreasing. Because the partial sum SkS_kS​k​​ can only increase as kkk becomes larger, the limit of the partial sum can either converge or diverge to ∞\infty∞. Therefore, it follows that a series ∑n=0∞an\sum_{n=0}^{\infty} a_n∑​n=0​∞​​a​n​​ with non-negative terms converges if and only if the sequence SkS_kS​k​​ of partial sums is bounded.

Example 1

The series ∑n≥11n2\sum_{n \ge 1} \frac{1}{n^2}∑​n≥1​​​n​2​​​​1​​ is convergent because of the inequality:

1n2≤1n−1−1n,(n≥2)\displaystyle{\frac1 {n^2} \le \frac{1}{n-1} - \frac{1}{n}, (n \ge 2)}​n​2​​​​1​​≤​n−1​​1​​−​n​​1​​,(n≥2) 

and because:

 ∑n≥2(1n−1−1n)=(1−12)+(12−13)+(13−14)+⋯=1\displaystyle{\sum_{n \ge 2} \left(\frac{1}{n-1} - \frac{1}{n} \right) =\left(1-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \cdots = 1}​n≥2​∑​​(​n−1​​1​​−​n​​1​​)=(1−​2​​1​​)+(​2​​1​​−​3​​1​​)+(​3​​1​​−​4​​1​​)+⋯=1

Example 2

Would the series

S=1+12+14+18+⋯+12n+⋯\displaystyle{S = 1 + \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots+ \frac{1}{2^n}+\cdots}S=1+​2​​1​​+​4​​1​​+​8​​1​​+⋯+​2​n​​​​1​​+⋯ 

converge? It is possible to "visualize" its convergence on the real number line? We can imagine a line of length 222, with successive segments marked off of lengths 111, 12\frac{1}{2}​2​​1​​, 14\frac{1}{4}​4​​1​​, etc. There is always room to mark the next segment, because the amount of line remaining is always the same as the last segment marked: when we have marked off 12\frac{1}{2}​2​​1​​, we still have a piece of length 12\frac{1}{2}​2​​1​​ unmarked, so we can certainly mark the next 14\frac{1}{4}​4​​1​​. This argument does not prove that the sum is equal to 222 (although it is), but it does prove that it is at most 222. In other words, the series has an upper bound. Proving that the series is equal to 222 requires only elementary algebra, however. If the series is denoted SSS, it can be seen that:

 $\displaystyle{\frac{S}{2} \,= \frac{1+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+\cdots}{2} \\ \quad= \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{16} +\cdots}$

Therefore:

$\displaystyle{S-\frac{S}{2} = 1\\S = 2}$

Geometric Sum

Visualization of the geometric sum in Example 2. The length of the line (222) can contain all the successive segments marked off of lengths 111, 12\frac{1}{2}​2​​1​​, 14\frac{1}{4}​4​​1​​, etc.

For these specific examples, there are easy ways to check the convergence. However, it could be the case that there are no easy ways to check the convergence. For these general cases, we can experiment with several well-known convergence tests (such as ratio test, integral test, etc.). We will learn some of these tests in the following atoms.

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