Series
Table of Contents
1 Definition
Given some (infinite) sequence \( (a_n)_n \) we can consider its infinite series, denoted:
\[ \sum_{n=1}^{\infty} a_n \]
We say the series converges iff the sequence of partial sums of the sequence converges.
\[ S_n = \sum_{i=1}^n a_i \]
If \( S_n \) converges we say the series is convergent, else we say the series is divergent.
2 Tests
There is no general process for checking if a series converges, but we can apply tests which provide definitive statements for some series.
2.1 Nth Term Test
If a series converges then its corresponding sequence must be null.
2.1.1 Proof
Suppose \( S_n \) converges to some \( L \) then by the shifted sequence rule, \( S_{n + 1} \) also converges to the same \( L \). By the algebra of limits theorem:
\[ (S_{n + 1} - S_n)_n \]
converges to \( L - L = 0 \), but
\[ (S_{n + 1} - S_n)_n = (a_n)_n \implies a_n \text{ converges to } 0 \]
2.2 Integral Test
Let the sequence \( (a_n)_n \) be defined by \( a_n = f(n) \) then if the following are true for \( f \):
- \( f \) is defined on the interval \( I = [1, \infty) \)
- \( f \) is monotonically decresing on \( I \)
- \( f \) is positive on \( I \)
Then the integral test states that:
\[ \sum_{n=1}^{\infty} a_n \text{ converges } \iff \int^{\infty}_{1} f(x) \ dx\text{ exists } \]
2.3 Comparison Test
Given two sequences \( a_n \) and \( b_n \), for which
- \( a_n, b_n > 0 \ \forall n \)
- \( a_n \le b_n \ \forall n \)
The comparison test states that:
\[ \sum_{n=1}^{\infty} b_n \text{ converges } \implies \sum_{n=1}^{\infty} a_n \text{ converges} \]
And
\[ \sum_{n=1}^{\infty} a_n \text{ diverges } \implies \sum_{n=1}^{\infty} b_n \text{ diverges} \]
2.3.1 Proof
Consider the sequence of partial sums: \( S_n = \sum_{k=1}^n a_k, \ R_n = \sum_{k=1}^n b_k \). Given the convergence of \( R_n \) we know that it must have an upper bound \( B \).
\[ \implies S_n \le R_n < B \ \forall n \] \[ \implies S_n < B \ \forall n \]
And we also have from (1) that \( S_n \) is monotone increasing. Hence since \( S_n \) is monotone increasing and bounded above, it must converge by the MCT.
2.4 Ratio Test
Let
\[ L = \frac{|a_{n + 1}|}{|a_n|} \]
Then the ratio test asserts that if:
- \( L < 1 \implies a_n \text{ converges absolutely} \)
- \( L > 1 \implies \text{ the series is divergent} \)
- And if \( L = 1 \) or the limit does not exist, then the test is inconclusive
The proof of this statement follows via construction of an appropriate geometric series and the comparison test.
2.5 Root Test
Let
\[ L = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}} \]
Then similar to the ratio test, the root test asserts that:
- \( L < 1 \implies a_n \text{ converges absolutely} \)
- \( L > 1 \implies \text{ the series is divergent} \)
- And if \( L = 1 \) or the limit does not exist, then the test is inconclusive
Note that if \( \lim_{n \to \infty } |a_n|^{\frac{1}{n}} \) exists then by the properties of the limit superior \( L = \lim_{n \to \infty } |a_n|^{\frac{1}{n}} \). We can also use this test to define the RoC as:
\[ R = \frac{1}{\limsup_{n \to \infty} |a_n|^{\frac{1}{n}}} \]
This works as the RoC makes no claims on the convergence at the boundaries, we can also substitute in \( \infty \) for the limit superior if \( (|a_n|^\frac{1}{n})_n \) does not have a convergent subsequence.
2.5.1 Proof
Suppose \( L < 1 \), then by the properties of the limit superior:
\[ \exists N \in \mathbb{N}, \ k \ s.t. \ |a_n|^\frac{1}{n} < k < 1, \ \forall n \ge N \]
Since \( f(x) = x^n \) is strictly increasing on the interval \( [0, \infty) \):
\[ \ |a_n| < k^n < 1, \ \forall n \ge N \]
Summing from \( N \) to \( \infty \):
\[ \sum_{n=N}^{\infty} |a_n| < \sum_{n=N}^{\infty} k^n \]
But \( \sum_{n=N}^{\infty} k^n \) is a convergent geometric series, hence \( \sum_{n=N}^{\infty} |a_n| \) converges by the comparison test.
\[ \implies \sum_{n=1}^{\infty} |a_n| \text{ converges} \]
Hence if property 1. above is satisfied, \( a_n \) not only converges, but converges absolutely. A similar argument can be applied to prove 3.
3 Problems
- Show \( \sum_{n=1}^{\infty} \frac{1}{n} \) does not converge and hence show that \( \sum_{n=1}^{\infty} \frac{1}{2n + 1} \) does not converge.