Limx-1-12-14-12n1-13-19-13n-

The correct option is A 4/3 Use G.P formula of summation ∞ Σ nbsp; p=l nbsp;=11 x Therefore, 11 1211 13=232=43 Hence, option 'A' is co ...

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Substitution Method to Remove Indeterminate Form

limx→∞1+12+14...

Question

$lim x \to \infty \frac{1 + \frac{1}{2} + \frac{1}{4} + . . . . + \frac{1}{2^{n}}}{1 + \frac{1}{3} + \frac{1}{9} + . . . . + \frac{1}{3^{n}}} =$

\frac{4}{3}

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\frac{3}{4}

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\frac{1}{2}

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0

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{lim}_{n \to \infty} \frac{1 + \frac{1}{2} + \frac{1}{4} + . . . + \frac{1}{2^{n}}}{1 + \frac{1}{3} + \frac{1}{9} + . . . + \frac{1}{3^{n}}}

lim n \to \infty \frac{1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n}}}{1 + \frac{1}{3} + \frac{1}{9} + \dots + \frac{1}{3^{n}}}

2 C_{0} + 2^{2} \frac{C_{1}}{2} + 2^{3} \frac{C_{2}}{3} + 2^{4} \frac{C_{3}}{4} + . . . . . . . . . + 2^{n + 1} \frac{C_{n}}{n + 1} = \frac{3^{n + 1} - 1}{n + 1}

Find the value of n such that $\frac{2}{3} (4 n - 1) - (2 n - \frac{1 + n}{3}) = \frac{1}{3} n + \frac{4}{3}$ .

n \sum r = 1 T_{r} = (3^{n} - 1),

n \sum r = 1 \frac{1}{T_{r}}

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