If \(\mathrm{a}_{\mathrm{n}}=\frac{3}{4}-\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{3}+\cdots(-1)^{\mathrm{n}-1}\left(\frac{3}{4}\right)^{\mathrm{n}}\) and \(\mathrm{b}_{\mathrm{n}}=1-\mathrm{a}_{\mathrm{n}}\), then find the minimum natural number \(\mathrm{n}_{0}\) such that \(\mathrm{b}_{\mathrm{n}}>\mathrm{a}_{\mathrm{n}} \forall \mathrm{n}>\mathrm{n}_{0}\)

JEE Advanced · 2006 · 09 Apr 2006 · Sequences and Series

a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots (- 1)^{n - 1} (\frac{3}{4})^{n}

and

b_{n} = 1 - a_{n}

, then find the minimum natural number

n_{0}

such that

b_{n} > a_{n} \forall n > n_{0}

Answer: 6

The series

a_{n}

is a geometric series with first term

\frac{3}{4}

and common ratio

- \frac{3}{4}

a_{n} = \frac{\frac{3}{4} ( 1 - ( - \frac{3}{4} ) ^{n} )}{1 + \frac{3}{4}} = \frac{3}{7} (1 - (- \frac{3}{4})^{n})

Given

b_{n} = 1 - a_{n}

, we need

b_{n} > a_{n}

1 - a_{n} > a_{n} \Rightarrow 2 a_{n} < 1 \Rightarrow a_{n} < \frac{1}{2}

\frac{3}{7} (1 - (- \frac{3}{4})^{n}) < \frac{1}{2}

1 - (- \frac{3}{4})^{n} < \frac{7}{6}

- (- \frac{3}{4})^{n} < \frac{1}{6}

(- \frac{3}{4})^{n} > - \frac{1}{6}

For even n:

(\frac{3}{4})^{n} < \frac{1}{6}

. Testing:

(\frac{3}{4})^{6} = \frac{729}{4096} \approx 0.178 > \frac{1}{6} \approx 0.167

, but

(\frac{3}{4})^{7} < \frac{1}{6}

.

For odd n:

- (\frac{3}{4})^{n} > - \frac{1}{6}

is satisfied for

n \geq 5

.

The minimum

n_{0}

such that the condition holds for all

n > n_{0}

is 6.

JEE Advanced · 2006 · 09 Apr 2006 · Numerical answer