Infinite Series

Definitions

Let {a_n} be a sequence. Then the infinite sum

\[\sum\limits_{n = 1}^\infty {{a_n}} = {a_1} + {a_2} + \ldots + {a_n} + \ldots \]

is called an infinite series, or, simply, series. The partial sums of the series are given by

\[\sum\limits_{n = 1}^n {{a_n}} = {a_1} + {a_2} + \ldots + {a_n},\]

where S_n is called the nth partial sum of the series. If the partial sums {S_n} converge to L as n → ∞, then we say that the infinite series converges to L:

\[\sum\limits_{n = 1}^\infty {{a_n}} = L,\;\; \text{if}\;\;\lim\limits_{n \to \infty } {S_n} = L.\]

Otherwise we say that the series \(\sum\limits_{n = 1}^\infty {{a_n}} \) diverges.

\(N\)th term test

If the series \(\sum\limits_{n = 1}^\infty {{a_n}} \) is convergent, then \(\lim\limits_{n \to \infty } {a_n} = 0.\)

Important!

The converse of this theorem is false. The convergence of \({{a_n}}\) to zero does not imply that the series \(\sum\limits_{n = 1}^\infty {{a_n}} \) converges. For example, the harmonic series \(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) diverges (see Example \(3\)), although \(\lim\limits_{n \to \infty } {a_n} = 0.\)

Equivalently, if \(\lim\limits_{n \to \infty } {a_n} \ne 0\) or this limit does not exist, then the series \(\sum\limits_{n = 1}^\infty {{a_n}} \) is divergent.

Properties of Convergent Series

Let \(\sum\limits_{n = 1}^\infty {{a_n}} = A \) and \(\sum\limits_{n = 1}^\infty {{b_n}} = B \) be convergent series and let \(c\) be a real number. Then

\(\sum\limits_{n = 1}^\infty {\left( {{a_n} + {b_n}} \right)} = A + B\)
\(\sum\limits_{n = 1}^\infty {c{a_n}} = cA\)

Solved Problems

Example 1.

Determine whether \[\sum\limits_{n = 1}^\infty {\sqrt[n]{3}}\] converges or diverges.

Solution.

Since \(\lim\limits_{n \to \infty } \sqrt[n]{3} = \lim\limits_{n \to \infty } {3^{\frac{1}{n}}} = 1,\) then the series \(\sum\limits_{n = 1}^\infty {\sqrt[n]{3}} \) diverges by the \(n\)th term test.

Example 2.

Investigate convergence of the series \[\sum\limits_{n = 1}^\infty {\frac{{{e^n}}}{{{n^2}}}}.\]

Solution.

Calculate the limit \(\lim\limits_{n \to \infty } {\frac{{{e^n}}}{{{n^2}}}}.\) Using L'Hopital's rule, we find

\[\lim\limits_{x \to \infty } \frac{{{e^x}}}{{{x^2}}} = \lim\limits_{x \to \infty } \frac{{{e^x}}}{{2x}} = \lim\limits_{x \to \infty } \frac{{{e^x}}}{2} = \infty .\]

Hence, the original series diverges by the \(n\)th term test.

See more problems on Page 2.

Calculus

Infinite Sequences and Series