4.8 Q-13 | Sequence And Series | Mathematics 11

Exercise 4.8 — Question 13

Problem Statement

Find the sum of the series to $n$ terms and to infinity (where applicable):

$1 + \frac{4}{3} + \frac{7}{9} + \frac{10}{27} + \dots$

Note: This is an arithmetic-geometric series where the numerators form an arithmetic sequence and the denominators form a geometric sequence.

Identifying the Structure

Write the general term by inspection:

Numerators: $1, 4, 7, 10, \dots$ — arithmetic sequence with $a = 1$ , $d = 3$
Denominators: $1, 3, 9, 27, \dots$ — geometric sequence with first term $1$ , common ratio $r = \frac{1}{3}$

So the general term is:

$T_{n} = \frac{1 + ( n - 1 ) \cdot 3}{3 ^{n - 1}} = \frac{3 n - 2}{3 ^{n - 1}} = (3 n - 2) (\frac{1}{3})^{n - 1}$

This is an arithmetic-geometric series with:

Arithmetic part: $a = 1$ , $d = 3$
Geometric ratio: $r = \frac{1}{3}$

Sum to $n$ Terms

Let: $S_{n} = 1 + \frac{4}{3} + \frac{7}{9} + \frac{10}{27} + \dots + \frac{3 n - 2}{3 ^{n - 1}}$

Multiply both sides by $r = \frac{1}{3}$ : $\frac{1}{3} S_{n} = \frac{1}{3} + \frac{4}{9} + \frac{7}{27} + \dots + \frac{3 n - 5}{3 ^{n - 1}} + \frac{3 n - 2}{3 ^{n}}$

Subtract the second equation from the first: $S_{n} - \frac{1}{3} S_{n} = 1 + (\frac{4 - 1}{3}) + (\frac{7 - 4}{9}) + \dots + (\frac{( 3 n - 2 ) - ( 3 n - 5 )}{3 ^{n - 1}}) - \frac{3 n - 2}{3 ^{n}}$

$\frac{2}{3} S_{n} = 1 + \frac{3}{3} + \frac{3}{9} + \dots + \frac{3}{3 ^{n - 1}} - \frac{3 n - 2}{3 ^{n}}$

$\frac{2}{3} S_{n} = 1 + 3 (\frac{\frac{1}{3} ( 1 - ( \frac{1}{3} ) ^{n - 1} )}{1 - \frac{1}{3}}) - \frac{3 n - 2}{3 ^{n}}$

The geometric sum in the middle (from $k = 1$ to $n - 1$ of $\frac{1}{3 ^{k}}$ ): $\sum_{k = 1}^{n - 1} \frac{1}{3 ^{k}} = \frac{\frac{1}{3} ( 1 - \frac{1}{3 ^{n - 1}} )}{\frac{2}{3}} = \frac{1}{2} (1 - \frac{1}{3 ^{n - 1}})$

So: $\frac{2}{3} S_{n} = 1 + 3 \cdot \frac{1}{2} (1 - \frac{1}{3 ^{n - 1}}) - \frac{3 n - 2}{3 ^{n}}$

$\frac{2}{3} S_{n} = 1 + \frac{3}{2} - \frac{3}{2 \cdot 3 ^{n - 1}} - \frac{3 n - 2}{3 ^{n}}$

$\frac{2}{3} S_{n} = \frac{5}{2} - \frac{9}{2 \cdot 3 ^{n}} - \frac{3 n - 2}{3 ^{n}}$

$\frac{2}{3} S_{n} = \frac{5}{2} - \frac{9 + 2 ( 3 n - 2 )}{2 \cdot 3 ^{n}}$

$\frac{2}{3} S_{n} = \frac{5}{2} - \frac{6 n + 5}{2 \cdot 3 ^{n}}$

Multiplying both sides by $\frac{3}{2}$ :

$S_{n} = \frac{15}{4} - \frac{6 n + 5}{4} \cdot \frac{1}{3 ^{n - 1}}$

Sum to Infinity

Since $∣ r ∣ = \frac{1}{3} < 1$ , the series converges.

As $n \to \infty$ , the term $\frac{6 n + 5}{4 \cdot 3 ^{n - 1}} \to 0$ (exponential decay dominates linear growth).

Therefore: $S_{\infty} = \frac{15}{4}$

Key Formulas Used

Formula	Expression
General term of AGS	$T_{n} = [a + (n - 1) d] r^{n - 1}$
Sum to infinity of AGS	$S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$ , $∥ r ∥ < 1$

Verification using direct formula: $S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}} = \frac{1}{1 - \frac{1}{3}} + \frac{3 \cdot \frac{1}{3}}{( 1 - \frac{1}{3} ) ^{2}} = \frac{3}{2} + \frac{1}{\frac{4}{9}} = \frac{3}{2} + \frac{9}{4} = \frac{6}{4} + \frac{9}{4} = \frac{15}{4} ✓$