Amazing limit

Calculus Level 4

lim n 1 ln ( 3 n 2 ) x = 1 n 1 3 x 2 = ? \large \lim_{n\to\infty} \frac1{\ln(3n-2)} \sum_{x=1}^n \frac1{3x-2} = \ ?

1 3 \frac {1}{3} 0 \infty 1

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1 solution

Satyajit Mohanty
Jul 20, 2015

Let ( x n ) n = 1 and ( y n ) n = 1 (x_n)_{n=1}^{\infty} \text{ and } (y_n)_{n=1}^{\infty} be two real sequences defined as

x i = i = 1 n 1 3 i 2 and y i = ln ( 3 i 2 ) x_i = \displaystyle \sum_{i=1}^{n} \dfrac{1}{3i-2} \text{ and } y_i = \ln(3i-2)

Now, ( y n ) n = 1 (y_n)_{n=1}^{\infty} is a strictly monotonic increasing function, and it diverges to \infty .

Therefore, by Stolz-Cesáro Theorem , we have:

lim n x n y n = lim n x n + 1 x n y n + 1 y n \lim_{n \to \infty} \frac{x_n}{y_n} = \lim_{n \to \infty} \frac{x_{n+1} - x_n}{y_{n+1} - y_n}

= lim n 1 3 n + 1 ln ( 3 n + 1 ) ln ( 3 n 2 ) =\lim_{n \to \infty} \dfrac{\frac{1}{3n+1}}{\ln(3n+1) - \ln(3n-2)}

= lim n 1 ( 3 n + 1 ) ln ( 1 + 3 3 n 2 ) =\lim_{n \to \infty} \dfrac{1}{(3n+1)\ln(1+\frac{3}{3n-2})}

= lim n 3 n 2 3 ( 3 n + 1 ) = 1 3 =\lim_{n \to \infty} \dfrac{3n-2}{3(3n+1)} = \boxed{\dfrac{1}{3}}


The last step is obtained from the pen-ultimate step as we know that lim x 0 ln ( 1 + x ) = x \lim_{x \to 0} \ln(1+x) = x

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