Q. 4 - 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.

$\frac{1}{301},\frac{8}{304},\frac{27}{309},\frac{64}{316},\frac{125}{325},\frac{216}{336},...$

Step-by-Step Solution

Answer

The given sequence is divergent.

Step 1. Given information.

Consider the given question,

$\frac{1}{301},\frac{8}{304},\frac{27}{309},\frac{64}{316},\frac{125}{325},\frac{216}{336},...$

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 $n\to \infty$,

$$\begin{gathered} \underset{n\to \infty }{\lim }\left\{a_n\right\}=\frac{n^3}{n^2+300} ; n\in N \\ =\underset{n\to \infty }{\lim }\frac{3n^2}{2n} \\ =\infty \end{gathered}$$

Thus, the given sequence not approaches a unique finite value.

Therefore, the given sequence diverges.