The Harmonic Series
One of the most important and surprising series in mathematics is the harmonic series:
This series is fascinating because it diverges, but it does so extremely slowly. The partial sums grow toward infinity, but at such a slow rate that it’s not immediately obvious the series diverges.
Look at how slowly the partial sums [latex]S_k[/latex] grow:
| [latex]k[/latex] | [latex]10[/latex] | [latex]100[/latex] | [latex]1000[/latex] | [latex]10,000[/latex] | [latex]100,000[/latex] | [latex]1,000,000[/latex] |
| [latex]{S}_{k}[/latex] | [latex]2.92897[/latex] | [latex]5.18738[/latex] | [latex]7.48547[/latex] | [latex]9.78761[/latex] | [latex]12.09015[/latex] | [latex]14.39273[/latex] |
Even after adding one million terms, the partial sum is only about [latex]14.4[/latex]. From this table alone, you might think the series converges to some finite value! Despite the slow growth, we can prove analytically that the harmonic series diverges by showing the partial sums are unbounded.
Step 1: Group terms strategically
Let’s examine the first few partial sums:
- [latex]S_1 = 1[/latex]
- [latex]S_2 = 1 + \frac{1}{2}[/latex]
- [latex]S_4 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}[/latex]
Step 2: Use inequalities to find lower bounds
For [latex]S_4[/latex], notice that: [latex]\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}[/latex]
Therefore, we conclude that: [latex]S_4 > 1 + \frac{1}{2} + \frac{1}{2} = 1 + 2 \cdot \frac{1}{2}[/latex]
Step 3: Extend the pattern
For [latex]S_8[/latex]: [latex]S_8 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}[/latex]
Group the terms: [latex]S_8 > 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right)[/latex]
[latex]S_8 > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1 + 3 \cdot \frac{1}{2}[/latex]
Step 4: Generalize the result
Following this pattern:
- [latex]S_1 = 1[/latex]
- [latex]S_2 = 1 + \frac{1}{2}[/latex]
- [latex]S_4 > 1 + 2 \cdot \frac{1}{2}[/latex]
- [latex]S_8 > 1 + 3 \cdot \frac{1}{2}[/latex]
In general: [latex]S_{2^j} > 1 + j \cdot \frac{1}{2}[/latex] for all [latex]j \geq 1[/latex]
Step 5: Conclude divergence
Since [latex]1 + j \cdot \frac{1}{2} \to \infty[/latex] as [latex]j \to \infty[/latex], the sequence [latex]{S_k}[/latex] is unbounded and therefore diverges.
harmonic series
The harmonic series [latex]\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots[/latex] diverges, even though its terms approach zero.
[latex]\\[/latex]
This series demonstrates that having [latex]\lim_{n \to \infty} a_n = 0[/latex] is necessary but not sufficient for series convergence. The harmonic series diverges so slowly that its partial sums grow approximately like [latex]\ln(n)[/latex], making it a borderline case between convergence and divergence.
Algebraic Properties of Convergent Series
Since infinite series are defined using limits of sequences, they inherit many algebraic properties from limits and sequences.
algebraic properties of convergent series
Let [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] be convergent series. Then:
- Sum Rule: The series [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)[/latex] converges and [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}+{b}_{n}\right)=\displaystyle\sum _{n=1}^{\infty }{a}_{n}+\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex].
- Difference Rule:The series [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}-{b}_{n}\right)[/latex] converges and [latex]\displaystyle\sum _{n=1}^{\infty }\left({a}_{n}-{b}_{n}\right)=\displaystyle\sum _{n=1}^{\infty }{a}_{n}-\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex].
- Constant Multiple Rule:For any real number [latex]c[/latex], the series [latex]\displaystyle\sum _{n=1}^{\infty }c{a}_{n}[/latex] converges and [latex]\displaystyle\sum _{n=1}^{\infty }c{a}_{n}=c\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex].
Evaluate