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Worked solution

Find rational sequence limits without a calculator

Without using a calculator, determine whether each sequence converges and, if it does, find the limit: (a) un=2n+1(3n−1)2u_n=\frac{2n+1}{(3n-1)^2}; (b) un=2n+13n−1u_n=\frac{2n+1}{3n-1}; (c) un=6n2+n+13n−1u_n=\frac{6n^2+n+1}{3n-1}.

CJC · Sequences and Series · Example 2

(a) Divide by the highest denominator power

Answer

Converges to 00.

Expand then divide numerator and denominator by n squared

2n+1(3n−1)2=2n+19n2−6n+1=(2n+1)/n2(9n2−6n+1)/n2=2/n+1/n29−6/n+1/n2→09=0.\begin{aligned}\frac{2n+1}{(3n-1)^2}&=\frac{2n+1}{9n^2-6n+1}\\&=\frac{(2n+1)/n^2}{(9n^2-6n+1)/n^2}\\&=\frac{2/n+1/n^2}{9-6/n+1/n^2}\to\frac09=0.\end{aligned}

This retains the complete Teacher substitution chain.

How did this part go?

(b) Two methods for the same limit

Answer

Converges to 2/32/3.

Method 1: long division

2n+13n−1=23+53(3n−1)→23\begin{aligned}&\frac{2n+1}{3n-1}\\ & =\frac23+\frac5{3(3n-1)}\to\frac23\end{aligned}

The remainder fraction tends to zero.

Method 2: divide throughout by n

2n+13n−1=(2n+1)/n(3n−1)/n=2+1/n3−1/n→23\begin{aligned}&\frac{2n+1}{3n-1}\\ & =\frac{(2n+1)/n}{(3n-1)/n}\\ & =\frac{2+1/n}{3-1/n}\to\frac23\end{aligned}

The denominator tends to 3, so the quotient limit law applies.

How did this part go?

(c) Expose the growing polynomial

Answer

Does not converge; tends to +∞+\infty.

Use long division

6n2+n+13n−1=2n+1+23n−1,23n−1→0,2n+1→+∞\frac{6n^2+n+1}{3n-1}=2n+1+\frac2{3n-1},\qquad\frac2{3n-1}\to0,\quad2n+1\to+\infty

The proper rational remainder vanishes while 2n+1 grows without bound.

How did this part go?

Common mistakes

Do not substitute infinity as if it were a real number. A vanishing remainder does not remove the growing polynomial part.

Solution notes

One Teacher conclusion for (b) is corrupted. The algebra, the second method and the Student conclusion agree that the limit is 2/3.