Q. 1 - Calculus

Question

If a sequence $a_1,a_2,a_3,...,a_k,...$ 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.

$\frac{1}{2},\frac{4}{4},\frac{9}{8},\frac{16}{16},\frac{25}{32},\frac{36}{64},...$

Step-by-Step Solution

Answer

The given sequence converges.

Step 1. Given information.

Consider the given question,

${{\left\{a_k\right\}}_0}^{\infty }=\frac{1}{2},\frac{4}{4},\frac{9}{8},\frac{16}{16},\frac{25}{32},\frac{36}{64},... ...... \left(i\right)$

Step 2. Calculate the sequence.

The given sequence is the combination of two sentences, the number sequence; $1,4,9,16,...k^2,...$ whose kth term is $k^2$ and a denominator sequence $2,4,8,16,32,...,2^k,...$ whose kth term is $2^k$.

On rewriting equation (i),

${{\left\{a_k\right\}}_0}^{\infty }=\frac{1}{2},\frac{4}{4},\frac{9}{8},\frac{16}{16},\frac{25}{32},\frac{36}{64},...,\frac{k^2}{2^k},...$

The kth term of the sequence is $a_k=\frac{k^2}{2^k}$.

Now,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left\{a_k\right\}=\underset{k\to \infty }{\lim }\left\{\frac{k^2}{2^k}\right\} \\ \underset{k\to \infty }{\lim }\left\{a_k\right\}=\underset{k\to \infty }{\lim }\left\{k^2\right\}·\left\{\frac{1}{2^{\infty }}\right\} \\ \underset{k\to \infty }{\lim }\left\{a_k\right\}=\underset{k\to \infty }{\lim }\left\{k^2\right\}·0 \\ \underset{k\to \infty }{\lim }\left\{a_k\right\}=0 \end{gathered}$$

Thus, $\underset{k\to \infty }{\lim }\left\{a_k\right\}=0$, then by definition of convergent sequence the given sequence converges.