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Sequences and Series — Diagnostic Tests

Unit Tests​

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

UT-1: Method of Differences — Telescoping Series​

Question:

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

(b) Hence find the sum ∑r=1n1r(r+1)\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)}.

(c) A student claims that ∑r=1n1r(r+2)\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)

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

1=A(r+1)+Br  ⟹  A=1,  A+B=0  ⟹  B=−11 = A(r+1) + Br \implies A = 1, \; A + B = 0 \implies B = -1

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

(b)

∑r=1n(1r−1r+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1=nn+1\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. 1r(r+2)\dfrac{1}{r(r+2)} does not telescope the same way because the gap between denominators is 2.

1r(r+2)=12(1r−1r+2)\frac{1}{r(r+2)} = \frac{1}{2}\left(\frac{1}{r} - \frac{1}{r+2}\right)

∑r=1n1r(r+2)=12[(1−13)+(12−14)+(13−15)+⋯+(1n−1n+2)]\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,…,n3, 4, \ldots, n cancel partially. The surviving terms are:

12(1+12−1n+1−1n+2)=12(32−2n+3(n+1)(n+2))=34−2n+32(n+1)(n+2)\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)}

UT-2: Binomial Expansion — Non-Integer Power Validity​

Question:

(a) Find the binomial expansion of (1+3x)−1/3(1 + 3x)^{-1/3} up to and including the term in x3x^3.

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

(c) Use the expansion to find an approximation for 11.033\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+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots with n=−13n = -\frac{1}{3} and replacing xx with 3x3x:

(1+3x)−1/3=1+(−13)(3x)+(−13)(−43)2(3x)2+(−13)(−43)(−73)6(3x)3(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+492⋅9x2+−28816⋅27x3= 1 - x + \frac{\frac{4}{9}}{2} \cdot 9x^2 + \frac{-\frac{28}{81}}{6} \cdot 27x^3

=1−x+2x2−2881⋅276x3=1−x+2x2−149x3= 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∣<1|3x| \lt 1, i.e., ∣x∣<13|x| \lt \dfrac{1}{3}.

(c) 11.033=(1+0.03)−1/3\dfrac{1}{\sqrt[3]{1.03}} = (1 + 0.03)^{-1/3}. Here x=0.03x = 0.03, which satisfies ∣x∣<13|x| \lt \frac{1}{3}.

(1+0.03)−1/3≈1−0.03+2(0.03)2−149(0.03)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)−149(0.000027)= 1 - 0.03 + 2(0.0009) - \frac{14}{9}(0.000027)

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

To 5 decimal places: 0.971760.97176.

Integration Tests​

Tests synthesis of sequences and series with other topics.

IT-1: Term-by-Term Differentiation of a Series (with Differentiation)​

Question:

(a) Differentiate both sides of the identity 11−x=∑n=0∞xn\displaystyle\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n for ∣x∣<1|x| \lt 1 with respect to xx, and hence find ∑n=1∞nxn−1\displaystyle\sum_{n=1}^{\infty} nx^{n-1}.

(b) Use your result to find the exact value of ∑n=1∞n2n\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: ddx ⁣(11−x)=1(1−x)2\dfrac{d}{dx}\!\left(\dfrac{1}{1-x}\right) = \dfrac{1}{(1-x)^2}.

Differentiating the RHS term-by-term: ddx ⁣(∑n=0∞xn)=∑n=1∞nxn−1\dfrac{d}{dx}\!\left(\sum_{n=0}^{\infty} x^n\right) = \sum_{n=1}^{\infty} nx^{n-1}.

Therefore:

∑n=1∞nxn−1=1(1−x)2for ∣x∣<1\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \quad \text{for } |x| \lt 1

(b) We need ∑n=1∞n2n=∑n=1∞n(12)n\displaystyle\sum_{n=1}^{\infty} \frac{n}{2^n} = \sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^n.

Note that ∑n=1∞nxn−1=1(1−x)2\displaystyle\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} with x=12x = \frac{1}{2}:

∑n=1∞n(12)n−1=1(1−12)2=11/4=4\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:

∑n=1∞n2n=12∑n=1∞n(12)n−1=12×4=2\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