Q. 2 - Calculus

Question

If a sequence $a_1,a_2,a_3,\dots ,a_k,\dots$ approaches a real-number limit as $k\to \infty$, then the sequence $\left\{a_k\right\}$converges. If the terms of the sequence do not get arbitrarily close to some real number, then the sequence diverges. Determine the general form $\left\{a_k\right\}$for each of the following sequences, and then use L’Hopital’s  rule to determine whether that sequence converges or diverges.

$\frac{\ln 1}{\ln 2},\frac{\ln 2}{\ln 3},\frac{\ln 3}{\ln 4},\frac{\ln 4}{\ln 5},\frac{\ln 5}{\ln 6},\frac{\ln 6}{\ln 7},...$

Step-by-Step Solution

Answer

The given sequence is convergent.

Step 1. Given information.

Consider the given question,

$\frac{\ln 1}{\ln 2},\frac{\ln 2}{\ln 3},\frac{\ln 3}{\ln 4},\frac{\ln 4}{\ln 5},\frac{\ln 5}{\ln 6},\frac{\ln 6}{\ln 7},...$

Step 2. Calculate the sequence.

A sequence say $\left\{a_n\right\}$ is said to be convergent if the nth term of the sequence approaches unique finite as n approaches $\infty$ otherwise sequence said to be divergent.

From the given sequence, nth term is $\left\{a_n\right\}=\frac{\ln n}{\ln n+1} ...... \left(i\right)$.

On evaluating limit of equation (i) as $n\to \infty$,

$$\begin{gathered} \underset{n\to \infty }{\lim }\left\{a_n\right\}=\underset{n\to =}{\lim }\frac{\ln n}{\ln n+1} \\ =\underset{n\to \infty }{\lim }\frac{1/n}{1/n+1} \\ =\underset{n\to \infty }{\lim }\frac{n+1}{n} \\ =\underset{n\to \infty }{\lim }\frac{n\left(1+\frac{1}{n}\right)}{n} \end{gathered}$$

Thus, $\underset{n\to \infty }{\lim }\left\{a_n\right\}=1 \left(\because \underset{n\to \infty }{\lim }\frac{1}{n}=0\right)$

Therefore, the given sequence is convergent.