Calculus

Infinite Sequences and Series

Sequences and Series Logo

Infinite Series

Definitions

Let \(\left\{ {{a_n}} \right\}\) 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 \(n\)th partial sum of the series. If the partial sums \(\left\{ {{S_n}} \right\}\) converge to \(L\) as \(n \to \infty,\) 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

Solved Problems

Click or tap a problem to see the solution.

Example 1

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

Example 2

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

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.

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