The value of lim _{n} {rightarrow infty}left[ | vedclass

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Question

(B) The given expression is $\lim _{n \rightarrow \infty} \sum_{k=1}^n \left(\frac{1}{2^k \cdot 3^k} + \frac{1}{2^{k+1} \cdot 3^k}\right)$.Factor out

Vedclass · Accepted Answer

$\frac{3}{10}$

Vedclass · Answer

(B) The given expression is $\lim _{n ightarrow \infty} \sum_{k=1}^n \left(\frac{1}{2^k \cdot 3^k} + \frac{1}{2^{k+1} \cdot 3^k}ight)$.Factor out the common terms in each bracket: $\frac{1}{2^k \cdot 3^k} (1 + \frac{1}{2}) = \frac{1}{6^k} \cdot \frac{3}{2}$.Thus, the sum becomes $\frac{3}{2} \sum_{k=1}^n \frac{1}{6^k}$.As $n ightarrow \infty$, this is an infinite geometric series with first term $a = \frac{1}{6}$ and common ratio $r = \frac{1}{6}$.The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{1/6}{1-1/6} = \frac{1/6}{5/6} = \frac{1}{5}$.Therefore, the total value is $\frac{3}{2} 	imes \frac{1}{5} = \frac{3}{10}$.

The value of $\lim _{n}$ ${\rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]$ is

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