If a_{n} = \frac{3}{4} - ( \frac{3}{4} )^{2} + ( \frac{3}{4} )^{3} + ( - 1)^{n - 1}( \frac{3}{4} )^{n} and b_{n} = 1 - a_{n}, then find the least natural number n_{0} such that b_{n} > a_{n}\forall n ? n_{0}: Sequence & Series (Mathematics)

\n\na_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots .+(-1)^{n-1}\left(\frac{3}{4}\right)^n\n\n

\n\nG.P. with common ratio

-\frac{1}{3}

\n\n

\n\n\begin{gathered}\n\na_n=\frac{\frac{3}{4}\left[1-\left(-\frac{3}{4}\right)^n\right]}{1-\left(-\frac{3}{4}\right)}=\frac{3}{7}\left[1-\left(-\frac{3}{4}\right)^n\right] \\\n\nb_n>a_n \Rightarrow 1-a_n>a_n \Rightarrow a_n<\frac{1}{2}\n\n\end{gathered}\n\n

\n \nFrom (1) and (2),

(-3)^{n+1}<2^{2 n-1}

\n\nIf

n

is even L.H.S. is negative, R.H.S. is positive and the inequality certainly holds.\n\nIf

n

is odd, the above holds for

n \geq 7

\n\n∴ The least natural number

n_0

for which it holds is 6 .\n

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