Method of differences and miscellaneous series

Most exam problems involve series that are neither AP nor GP, but can still be summed in closed form by clever tricks. The two key ideas are: (1) method of differences , when consecutive differences form a known progression; and (2) telescoping , when the general term can be written as $f(k+1) - f(k)$ so the sum collapses.

Method of differences

Suppose $a_1, a_2, a_3, \dots$ is a sequence whose differences $d_k = a_{k+1} - a_k$ form a simpler progression (often AP or GP). Then $a_n = a_1 + \sum_{k=1}^{n-1} d_k.$ You sum the differences (which is easier), then add to $a_1$.

Example pattern. $4, 11, 22, 37, 56, \dots$ Differences: $7, 11, 15, 19, \dots$ , an AP. So $d_k = 7 + 4(k - 1) = 4k + 3$. Then $a_n = 4 + \sum_{k=1}^{n-1} (4k + 3) = 4 + 4 \cdot \frac{(n-1)n}{2} + 3(n - 1) = 2n^2 + n + 1.$

Telescoping

If you can write $T_k = f(k+1) - f(k)$ (or similar), then $\sum_{k=1}^n T_k = f(n + 1) - f(1).$ Most middle terms cancel.

Standard fact. $\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$. So $\sum_{k=1}^n \frac{1}{k(k+1)} = 1 - \frac{1}{n+1} = \frac{n}{n+1}.$

More generally:

• $\dfrac{1}{k(k+1)(k+2)} = \dfrac{1}{2}\left[\dfrac{1}{k(k+1)} - \dfrac{1}{(k+1)(k+2)}\right]$.
• $\dfrac{1}{(2k-1)(2k+1)} = \dfrac{1}{2}\left[\dfrac{1}{2k-1} - \dfrac{1}{2k+1}\right]$.

Arithmetic-Geometric Series (AGP)

A series like $1 + 2x + 3 x^2 + 4 x^3 + \dots$ , each term is an AP times a GP. Sum of $n$ terms (for $r \ne 1$): $S_n = \sum_{k=1}^n (a + (k-1)d) r^{k-1}.$ Trick. Compute $S_n - r S_n$, regroup, sum the GP that emerges. Result: $S_n = \frac{a}{1 - r} + \frac{d \cdot r (1 - r^{n-1})}{(1 - r)^2} - \frac{[a + (n-1)d] r^n}{1 - r}.$

For $|r| < 1$ and $n \to \infty$: $S_\infty = \frac{a}{1 - r} + \frac{d r}{(1 - r)^2}.$

Worked examples

Example 1. Find the $n$-th term and the sum of $3, 8, 15, 24, 35, \dots$

Differences: $5, 7, 9, 11, \dots$ , AP with $a = 5, d = 2$. So $T_n = 3 + \sum_{k=1}^{n-1}(2k + 3) = 3 + (n-1)n + 3(n-1) = n^2 + 2n.$ Sum: $\sum_{k=1}^n (k^2 + 2k) = \dfrac{n(n+1)(2n+1)}{6} + n(n+1)$.

Example 2. Sum $\dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dots + \dfrac{1}{n(n+1)}$.

$\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1}$. Sum telescopes to $1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$.

Example 3. Sum $\dfrac{1}{1 \cdot 3} + \dfrac{1}{3 \cdot 5} + \dfrac{1}{5 \cdot 7} + \dots$ to $n$ terms.

$\dfrac{1}{(2k-1)(2k+1)} = \dfrac{1}{2}\left[\dfrac{1}{2k-1} - \dfrac{1}{2k+1}\right]$. Sum telescopes to $\dfrac{1}{2}\left[1 - \dfrac{1}{2n+1}\right] = \dfrac{n}{2n+1}$.

Example 4. Sum the AGP $1 + 2x + 3x^2 + 4x^3 + \dots + n x^{n-1}$.

This is the derivative of $1 + x + x^2 + \dots + x^n$. Or use the AGP method: $S = 1 + 2x + 3x^2 + \dots + n x^{n-1}$ $x S = x + 2x^2 + 3x^3 + \dots + n x^n$ $S - x S = 1 + x + x^2 + \dots + x^{n-1} - n x^n = \dfrac{1 - x^n}{1 - x} - n x^n$. So $S = \dfrac{1 - (n+1) x^n + n x^{n+1}}{(1 - x)^2}$.

For $|x| < 1, n \to \infty$: $S = \dfrac{1}{(1 - x)^2}$.

Example 5. Find the sum to $n$ terms of $1 \cdot 2^2 + 2 \cdot 3^2 + 3 \cdot 4^2 + \dots$

$T_k = k(k + 1)^2 = k(k^2 + 2k + 1) = k^3 + 2k^2 + k$. So $\sum_{k=1}^n T_k = \sum k^3 + 2 \sum k^2 + \sum k = \left(\frac{n(n+1)}{2}\right)^2 + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}.$

Combine over common factor $\dfrac{n(n+1)}{12}$: $\dfrac{n(n+1)(3 n^2 + 11 n + 10)}{12} = \dfrac{n(n+1)(n+2)(3n + 5)}{12}$.

Try it yourself

1. Find the $n$-th term of $2, 5, 10, 17, 26, \dots$
2. Sum to $n$ terms: $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots$
3. Sum: $\dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dots$ to $n$ terms.
4. Sum the AGP $1 + 3x + 5x^2 + 7x^3 + \dots$ to $n$ terms.
5. Find the $n$-th term of $1, 5, 14, 30, 55, \dots$ (Hint: it equals $\dfrac{n(n+1)(2n+1)}{6}$.)
6. Sum to $\infty$: $1 + \dfrac{3}{2} + \dfrac{5}{4} + \dfrac{7}{8} + \dots$
7. Sum to $n$ terms: $\dfrac{1}{2 \cdot 5} + \dfrac{1}{5 \cdot 8} + \dfrac{1}{8 \cdot 11} + \dots$
8. Find the sum to $n$ terms of $4 + 44 + 444 + \dots$ (Hint: $\dfrac{4}{9}(10^k - 1)$.)
9. Find the $n$-th term and sum of $3, 7, 13, 21, 31, \dots$
10. Sum: $1 + (1 + 2) + (1 + 2 + 3) + \dots$ to $n$ terms.
11. Sum: $1^2 + 3^2 + 5^2 + \dots$ to $n$ terms.
12. Sum to $\infty$: $\sum_{k=1}^{\infty} \dfrac{k}{2^k}$ (AGP with $a = 0, d = 1, r = 1/2$).

Pitfalls / Tricks

• When stuck on a non-AP non-GP series, take differences. If they are familiar, you have a route.
• For fractions like $\dfrac{1}{k(k+m)}$, decompose into partial fractions and telescope.
• For AGP, multiply by the common ratio and subtract.
• Insight. The trick is always to convert the given series into a sum of standard series (AP, GP, or one of the three power sums). Once you write $T_k$ in those terms, you're done.