Sequences and Series -- Diagnostic Tests

Sequences are like mathematical recipes — each term follows a rule, and series are the cumulative sum of following that recipe step by step: Arithmetic and geometric sequences model different types of growth — constant addition versus constant multiplication — capturing patterns from savings accounts to population dynamics

Why it matters: Sequences and series underpin financial mathematics, computer algorithms, and modeling natural phenomena

The key insight: Arithmetic and geometric sequences model different types of growth — constant addition versus constant multiplication — capturing patterns from savings accounts to population dynamics

Tests edge cases, boundary conditions, and common misconceptions for sequences and series.

Question:

(a) Express $\dfrac{1}{r(r+1)}$ in partial fractions.

(b) Hence find the sum $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}$.

(c) A student claims that $\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+2)}$ telescopes in the same way. Determine whether this is true, and if not, find the correct sum.

[Difficulty: hard. Tests telescoping series with non-adjacent denominators, a common stumbling block.]

Solution:

(a)

$\frac{1}{r(r+1)} = \frac{A}{r} + \frac{B}{r+1}$

$1 = A(r+1) + Br \implies A = 1, \; A + B = 0 \implies B = -1$

$\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$

(b)

$\sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$

(c) The student is incorrect. $\dfrac{1}{r(r+2)}$ does not telescope the same way because the gap between denominators is 2.

$\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)$

$\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+2}\right)\right]$

Terms with denominators $3, 4, \ldots, n$ cancel partially. The surviving terms are:

$\frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{2}\left(\frac{3}{2} - \frac{2n+3}{(n+1)(n+2)}\right) = \frac{3}{4} - \frac{2n+3}{2(n+1)(n+2)}$

Question:

(a) Find the binomial expansion of $(1 + 3x)^{-1/3}$ up to and including the term in $x^3$.

(b) State the range of values of $x$ for which the expansion is valid.

(c) Use the expansion to find an approximation for $\dfrac{1}{\sqrt[3]{1.03}}$Giving your answer to 5 decimal places.

[Difficulty: hard. Tests binomial expansion with fractional exponent, validity range, and numerical application.]

Solution:

(a) Using $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$ with $n = -\frac{1}{3}$ and replacing $x$ with $3x$:

$(1 + 3x)^{-1/3} = 1 + \left(-\frac{1}{3}\right)(3x) + \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2}(3x)^2 + \frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\left(-\frac{7}{3}\right)}{6}(3x)^3$

$= 1 - x + \frac{\frac{4}{9}}{2} \cdot 9x^2 + \frac{-\frac{28}{81}}{6} \cdot 27x^3$

$= 1 - x + 2x^2 - \frac{28}{81} \cdot \frac{27}{6}x^3 = 1 - x + 2x^2 - \frac{14}{9}x^3$

(b) The expansion is valid when $|3x| \lt 1$I.e., $|x| \lt \dfrac{1}{3}$.

(c) $\dfrac{1}{\sqrt[3]{1.03}} = (1 + 0.03)^{-1/3}$. Here $x = 0.03$Which satisfies $|x| \lt \frac{1}{3}$.

$(1 + 0.03)^{-1/3} \approx 1 - 0.03 + 2(0.03)^2 - \frac{14}{9}(0.03)^3$

$= 1 - 0.03 + 2(0.0009) - \frac{14}{9}(0.000027)$

$= 1 - 0.03 + 0.0018 - 0.000042 = 0.971758$

To 5 decimal places: $0.97176$.

Tests synthesis of sequences and series with other topics.

Question:

(a) Differentiate both sides of the identity $\displaystyle\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| \lt 1$ with respect to $x$ And hence find $\displaystyle\sum_{n=1}^{\infty} nx^{n-1}$.

(b) Use your result to find the exact value of $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n}$.

[Difficulty: hard. Combines infinite series with differentiation to derive new summation formulas.]

Solution:

(a) Differentiating the LHS: $\dfrac{d}{dx}\!\left(\dfrac{1}{1-x}\right) = \dfrac{1}{(1-x)^2}$.

Differentiating the RHS term-by-term: $\dfrac{d}{dx}\!\left(\sum_{n=0}^{\infty} x^n\right) = \sum_{n=1}^{\infty} nx^{n-1}$.

Therefore:

$\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \quad \text{for } |x| \lt 1$

(b) We need $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = \sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^n$.

Note that $\displaystyle\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$ with $x = \frac{1}{2}$:

$\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n-1} = \frac{1}{\left(1 - \frac{1}{2}\right)^2} = \frac{1}{1/4} = 4$

Our target sum is:

$\sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{1}{2}\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n-1} = \frac{1}{2} \times 4 = 2$

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