Product

Calculus Level 5

$\lim_{n \rightarrow \infty} \left( \frac{N^{2}+N+1}{3} \right) \left( \frac{N^{6}+N^{3}+1}{3} \right) \left( \frac{N^{18}+N^{9}+1}{3} \right) \ldots \left( \frac{N^{2 \cdot 3^{n}}+N^{3^{n}}+1}{3} \right)$

For integer $n$ , let $N = 1+\frac{1}{3^{n}}$ . Find the value of the limit above to nearest integer.

The answer is 6.

1 solution

คลุง แจ็ค
Sep 17, 2019

Consider

$\displaystyle N^{2\cdot 3^{n}}+N^{3^{n}}+1=\dfrac{N^{3^{n+1}}-1}{N^{3^{n}}-1}$

So, the product is somewhat telescopic and nearly all the terms cancel and we get,

$\displaystyle \dfrac{N^{3^{n+1}}-1}{N-1} \times \dfrac{1}{3^{n+1}}$

Substitute an expression for N, we have,

$\displaystyle \dfrac{(1+\dfrac{1}{3^{n}})^{3^{n+1}}-1}{3}$

which tends to

$\displaystyle \dfrac{e^{3}-1}{3} \approx 6.3618$

So, the nearest integer is therefore $\displaystyle\boxed{6}$

Product

The answer is 6.

1 solution

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