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Harmonic and Special Series
Harmonic Progression (HP)
A sequence is a Harmonic Progression if the reciprocals of its terms form an Arithmetic Progression. The general form is $\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \dots$
$$H_n = \frac{1}{a + (n-1)d}$$
The nth term of an HP is the reciprocal of the nth term of the corresponding AP
$H_n$=nth term of the HP
$a$=First term of the corresponding AP
$d$=Common difference of the corresponding AP
$n$=Position of the term
$a + (n-1)d = 0$
→The term is undefined — HP cannot contain zero
Reciprocal Link: To solve any HP problem, convert terms to their reciprocals to form an AP, solve in the AP domain, then convert back.
No Zero Rule: Since $1/0$ is undefined, zero can never be a term of an HP.
No Direct Formula: There is no direct formula for the sum of an HP — always convert to AP first.
Harmonic Mean (HM)
The Harmonic Mean $H$ between two numbers $a$ and $b$ is defined so that $a, H, b$ are in HP. Equivalently, $\frac{1}{a}, \frac{1}{H}, \frac{1}{b}$ must be in AP.
$$H = \frac{2ab}{a+b}$$
The HM is the reciprocal of the arithmetic mean of the reciprocals
$H$=Harmonic Mean between $a$ and $b$
$a, b$=The two numbers
$a = b$
→$H = a = b$ (all three means coincide)
$a + b = 0$
→$H$ is undefined (division by zero)
Derivation: Since $\frac{1}{a}, \frac{1}{H}, \frac{1}{b}$ are in AP, $\frac{1}{H} - \frac{1}{a} = \frac{1}{b} - \frac{1}{H}$, which gives $\frac{2}{H} = \frac{1}{a} + \frac{1}{b}$.
Verification Shortcut: For HM between 3 and 7: $H = \frac{2(3)(7)}{3+7} = \frac{42}{10} = \frac{21}{5}$. Check: $\frac{1}{3}, \frac{5}{21}, \frac{1}{7}$ has common difference $\frac{5}{21} - \frac{1}{3} = \frac{-2}{21}$ and $\frac{1}{7} - \frac{5}{21} = \frac{-2}{21}$.
To insert $n$ Harmonic Means between two numbers $a$ and $b$, first insert $n$ AMs between $\frac{1}{a}$ and $\frac{1}{b}$, then take their reciprocals.
$$d = \frac{\frac{1}{b} - \frac{1}{a}}{n+1} = \frac{a-b}{ab(n+1)}$$
The common difference of the AP formed by $\frac{1}{a}, A_1, A_2, \dots, A_n, \frac{1}{b}$
$d$=Common difference of the reciprocal AP
$n$=Number of HMs to insert
$a, b$=The boundary terms of the HP
Step-by-Step Method: (1) Take reciprocals $\frac{1}{a}$ and $\frac{1}{b}$. (2) Find $d = \frac{\frac{1}{b} - \frac{1}{a}}{n+1}$. (3) The AMs are $\frac{1}{a} + d, \frac{1}{a} + 2d, \dots$ (4) Take reciprocals of these AMs.
General $k$th HM: The $k$th harmonic mean is $\frac{ab(n+1)}{(n+1-k)b + ka}$.
Relations between AM, GM, and HM
For any two numbers $a$ and $b$, the Arithmetic Mean $A$, Geometric Mean $G$, and Harmonic Mean $H$ satisfy a fundamental identity that links all three.
$$G^2 = A \cdot H$$
The three Pythagorean means are connected — A, G, H are always in GP
$A$=$\frac{a+b}{2}$
$G$=$\pm\sqrt{ab}$
$H$=$\frac{2ab}{a+b}$
GP Connection: Since $G^2 = AH$, the means $A, G, H$ are themselves in Geometric Progression.
Universal: The identity $G^2 = AH$ holds even when $a, b$ are complex numbers.
Finding Numbers: If any two of $A$, $G$, $H$ are known, the two numbers $a$ and $b$ satisfy $a + b = 2A$ and $ab = G^2 = AH$, so $a$ and $b$ are roots of $x^2 - 2Ax + AH = 0$.
For any two distinct positive real numbers, the three means always satisfy a strict ordering.
$$A > G > H \quad (a, b > 0, \; a \neq b)$$
AM is always the largest and HM the smallest for distinct positive reals
$A > G$=Equivalent to $(\sqrt{a} - \sqrt{b})^2 > 0$
$G > H$=Equivalent to $a + b > 2\sqrt{ab}$, same inequality
Proof Sketch for $A > G$: $\frac{a+b}{2} > \sqrt{ab} \Leftrightarrow a + b - 2\sqrt{ab} > 0 \Leftrightarrow (\sqrt{a} - \sqrt{b})^2 > 0$, which is true for distinct positive reals.
Equality Case: $A = G = H$ if and only if $a = b$ (all means coincide for equal numbers).
Negative Numbers: For two distinct negative reals with $G = -\sqrt{ab}$, the ordering reverses: $A < G < H$.
Verification Strategy: Given $a$ and $b$, compute all three means and check the ordering. For $a = 2, b = 8$: $A = 5$, $G = 4$, $H = \frac{16}{5} = 3.2$, confirming $5 > 4 > 3.2$.
Sigma Notation and Summation Properties
The Greek letter $\sum$ (sigma) provides a compact way to express sums. The notation $\sum_{k=m}^{n} a_k$ means $a_m + a_{m+1} + \dots + a_n$.
$$\sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n$$
A compact notation for expressing the sum of a sequence of terms
$k$=Index of summation (dummy variable)
$m, n$=Lower and upper limits of summation
$a_k$=General term expressed as a function of $k$
Linearity: $\sum (ca_k + b_k) = c\sum a_k + \sum b_k$ — constants factor out, sums split.
Constant Sum: $\sum_{k=1}^{n} c = nc$ — a constant summed $n$ times is $nc$.
Dummy Variable: The index letter is arbitrary — $\sum_{k=1}^{n} k^2 = \sum_{j=1}^{n} j^2 = \sum_{i=1}^{n} i^2$.
Three fundamental summation formulas for powers of natural numbers are derived using the telescoping identity $\sum_{k=1}^{n}[k^m - (k-1)^m] = n^m$.
$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$
Closed-form expression for the sum of squares of the first n natural numbers
$\sum k^2$=Sum of squares: $1^2 + 2^2 + \dots + n^2$
$n$=Number of terms
Sum of $n$: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ — the triangular numbers.
Sum of Cubes: $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$ — exactly the square of the sum of $n$.
Telescoping Derivation: Setting $m = 2$ in $\sum[k^2 - (k-1)^2] = n^2$ gives $\sum(2k - 1) = n^2$, which leads to $2\sum k - n = n^2$, so $\sum k = \frac{n(n+1)}{2}$.
Quick Check: For $n = 3$: $\sum k = 6$ vs $\frac{3(4)}{2} = 6$; $\sum k^2 = 14$ vs $\frac{3(4)(7)}{6} = 14$; $\sum k^3 = 36$ vs $6^2 = 36$.
Telescoping Series (Method of Differences)
The Method of Differences simplifies a series by expressing each term $T_r$ as a difference $V_r - V_{r-1}$, causing intermediate terms to cancel.
$$T_r = V_r - V_{r-1} \implies S_n = V_n - V_0$$
The sum collapses to just the first and last values of the decomposition function
$T_r$=General term of the original series
$V_r$=The decomposition function chosen so that $T_r = V_r - V_{r-1}$
$S_n$=Sum of first $n$ terms
Cancellation Mechanism: $(V_1 - V_0) + (V_2 - V_1) + (V_3 - V_2) + \dots + (V_n - V_{n-1}) = V_n - V_0$.
Partial Fractions: The key technique — decompose $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$, so $V_r = -\frac{1}{r+1}$.
Limiting Case: As $n \to \infty$, if $V_n \to L$, then $S_\infty = L - V_0$. For $\sum \frac{1}{r(r+1)}$: $S_\infty = 0 - (-1) = 1$.