Question

How do you determine whether the sequence ${{a}_{n}}=\sqrt{{{n}^{2}}+n}-n$ converges, if so how do you find the limit?

Answer 1

We are given a sequence ${{a}_{n}}=\sqrt{{{n}^{2}}+n}-n$ and we have to check if the given sequence converges and if it converges, we have to also find the limit. We will be using the root test to check the convergence of the given sequence which is as follows, $\displaystyle \lim_{n \to \infty }\sqrt[n]{|{{a}_{n}}|}=L$. Here, $L$ is the limit and if $L<1$ then the sequence converges, if $L>1$ the sequence diverges and if $L=1$ it is inconclusive. And hence we will have the answer to the question.

Complete step-by-step answer:
According to the given question, we are given a sequence and we are asked to check if the given sequence converges or not and if it does so, at what limit does the sequence converge.
The given sequence we have is,
${{a}_{n}}=\sqrt{{{n}^{2}}+n}-n$----(1)
We will use the root test to check if the sequence converges or not and the expression is written as,
$\displaystyle \lim_{n \to \infty }\sqrt[n]{|{{a}_{n}}|}=L$
Here, $L$ is the limit and if $L<1$ then the sequence converges, if $L>1$ the sequence diverges and if $L=1$ it is inconclusive, that is, the sequence can either converge or diverge.
If we directly apply the test on the sequence given in equation (1), we will get a value in indeterminate form, so we will first reduce the sequence as much as possible.
We will now multiply and divide the sequence by its conjugate, we will get,
$\Rightarrow {{a}_{n}}=\sqrt{{{n}^{2}}+n}-n\times \dfrac{\sqrt{{{n}^{2}}+n}+n}{\sqrt{{{n}^{2}}+n}+n}$
Solving the above expression, we have,
$\Rightarrow {{a}_{n}}=\dfrac{{{\left( \sqrt{{{n}^{2}}+n} \right)}^{2}}-{{n}^{2}}}{\sqrt{{{n}^{2}}+n}+n}$
Opening up the brackets, we will get,
$\Rightarrow {{a}_{n}}=\dfrac{{{n}^{2}}+n-{{n}^{2}}}{\sqrt{{{n}^{2}}+n}+n}$
$\Rightarrow {{a}_{n}}=\dfrac{n}{\sqrt{{{n}^{2}}+n}+n}$
We will now take ‘n’ common in the denominator and we get,
$\Rightarrow {{a}_{n}}=\dfrac{n}{n\left( \sqrt{1+\dfrac{1}{n}}+1 \right)}$
$\Rightarrow {{a}_{n}}=\dfrac{1}{\sqrt{1+\dfrac{1}{n}}+1}$-----(2)
Now, we will use the root test, we have,
$\displaystyle \lim_{n \to \infty }{{a}_{n}}$
$\Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{1}{\sqrt{1+\dfrac{1}{n}}+1}$
Applying the limit, we get,
$\Rightarrow \dfrac{1}{\sqrt{1+\dfrac{1}{\infty }}+1}$
$\Rightarrow \dfrac{1}{\sqrt{1+0}+1}$
$\Rightarrow \dfrac{1}{1+1}$
$\Rightarrow \dfrac{1}{2}$
That is, we get the value of the sequence as $\dfrac{1}{2}$.
As the value we got $\dfrac{1}{2}$ is less than 1,
Therefore, the given sequence converges and it converges at $\dfrac{1}{2}$.

Note: The sequence should be simplified if on applying the limit you get the indeterminate form. Also, the expression should be carefully simplified and the values substituted should be done in steps.