Question

Limits of sequences: Determine whether the sequences that follow are bounded, monotonic and/or eventually monotonic.

Determine whether each sequence converges or diverges. If the sequence converges, find its limit.

$\left\{\frac{k^2}{k+3}-\frac{k^2}{k+4}\right\}$

Short answer

• The sequence is eventually a monotonically increasing sequence.
• The bounds of the sequence are 0 and 1.
• The limit of the sequence is 1.

Step-by-step solution

Step 1. Given Information

The given sequence is$\left\{\frac{k^2}{k+3}-\frac{k^2}{k+4}\right\}$.

Step 2. Simplify the Expression

Simplify the expression as follows:

$\begin{array}{rcl}\frac{k^2}{k+3}-\frac{k^2}{k+4}& =& \frac{k^3+4k^2-k^3-3k^2}{(k+3)(k+4)}\\ & =& \frac{k^2}{(k+3)(k+4)}\end{array}$

Step 3. Check for monotonicity

• Subtract $a_k$from $a_{k+1}$.

role="math" localid="1649663735792" $\begin{array}{rcl}{a}_{k+1}-a_k& =& \frac{{(k+1)}^2}{(k+4)(k+5)}-\frac{k^2}{(k+3)(k+4)}\\ & =& \frac{{(k+1)}^2(k+3)-k^2(k+5)}{(k+3)(k+4)(k+5)}\\ & =& \frac{k^3+5k^2+7k+3-k^3-5k^2}{(k+3)(k+4)(k+5)}\\ & =& \frac{7k+3}{(k+3)(k+4)(k+5)}\end{array}$

• The obtained expression is positive for all values of k.
• So, the sequence is monotonically increasing.

Step 4. Check Boundedness and Limit

• The calculation of the previous step suggests the minimum value of the sequence is at $k=1$.
• So, the lower bound is $\begin{array}{rcl}\frac{1^2}{1+3}-\frac{1^2}{1+4}& =& \frac{1}{3}-\frac{1}{4}\\ & =& \frac{1}{12}\end{array}$
• The simplified expression of the general term of the sequence has same degree in the numerator and denominator.
• So, the limit is calculated as follows:

$\begin{array}{rcl}\underset{k\to \infty }{\lim }\frac{k^2}{(k+3)(k+4)}& =& \underset{k\to \infty }{\lim }\frac{\frac{k^2}{k^2}}{{\displaystyle \frac{k^2}{k^2}}+{\displaystyle \frac{7k}{k^2}}+{\displaystyle \frac{12}{k^2}}}\\ & =& \underset{k\to \infty }{\lim }\frac{1}{1+{\displaystyle \frac{7}{k}}+{\displaystyle \frac{12}{k^2}}}\\ & =& 1\end{array}$