Question

How do you test the series $\sum {\dfrac{{\sqrt {n + 2} - \sqrt n }}{n}}$from $n$ is $\left[ {1,\infty } \right)$ for convergence?

Answer 1

Hint : In order to test the above given series for convergence let the term written inside the summation as ${a_n}$ . Apply the operation of rationalise on this term and simplify it further using identity $\left( {A - B} \right)\left( {A + B} \right) = {A^2} - {B^2}$ . Use the p-series test which states $\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^p}}}}$ is convergent for $p > 1$ and divergent otherwise to obtain the given series is convergent.

Complete step-by-step answer :
We are given a summation series of the form $\sum {\dfrac{{\sqrt {n + 2} - \sqrt n }}{n}}$from $n$ is $\left[ {1,\infty } \right)$ .
We can rewrite the summation by including the limits in the summation symbol as
$\sum\limits_{n = 1}^\infty {\dfrac{{\sqrt {n + 2} - \sqrt n }}{n}}$
let ${a_n} = \dfrac{{\sqrt {n + 2} - \sqrt n }}{n}$ ,
Now let’s rationalise the ${a_n}$ term by multiplying and dividing ${a_n}$ with $\sqrt {n + 2} + \sqrt n$ , we get
${a_n} = \dfrac{{\sqrt {n + 2} - \sqrt n }}{n} \\\ {a_n} = \left( {\dfrac{{\sqrt {n + 2} - \sqrt n }}{n}} \right) \times \left( {\dfrac{{\sqrt {n + 2} + \sqrt n }}{{\sqrt {n + 2} + \sqrt n }}} \right) \;$
Simplifying further using the identity $\left( {A - B} \right)\left( {A + B} \right) = {A^2} - {B^2}$ by considering A as $\sqrt {n + 2}$ and B as $\sqrt n$ , we have
${a_n} = \dfrac{{{{\left( {\sqrt {n + 2} } \right)}^2} - {{\left( {\sqrt n } \right)}^2}}}{{n\left( {\sqrt {n + 2} + \sqrt n } \right)}} \\\ {a_n} = \dfrac{{n + 2 - n}}{{n\left( {\sqrt {n + 2} + \sqrt n } \right)}} \;$
${a_n} = \dfrac{2}{{n\left( {\sqrt {n + 2} + \sqrt n } \right)}}$
Now as you can see if we decrease the denominator , we will get a sequence which is greater. So if we remove 2 from the denominator, we get

$${a_n} = \dfrac{2}{{n\left( {\sqrt {n + 2} + \sqrt n } \right)}} < \dfrac{2}{{n\left( {\sqrt n + \sqrt n } \right)}} = \dfrac{2}{{2n\sqrt n }} \\\ {a_n} = \dfrac{2}{{2n\sqrt n }} \\\$$

Simplifying further using the property of exponents as ${a^m} \times {a^n} = {a^{m + n}}$ ,we have

$${a_n} = \dfrac{1}{{{n^{1 + \dfrac{1}{2}}}}} \\\ {a_n} = \dfrac{1}{{{n^{\dfrac{3}{2}}}}} \;$$

As we know that the series $\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^{\dfrac{3}{2}}}}}}$ is always convergent according to the P-series test which says that the series of the form $\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^p}}}}$ is convergent for $p > 1$ and divergent otherwise . So in our case we have $p = \dfrac{3}{2} = 1.5 > 1$ .
Therefore, we can say that the series $\sum\limits_{n = 1}^\infty {\dfrac{{\sqrt {n + 2} - \sqrt n }}{n}}$ is convergent from $n$ is $\left[ {1,\infty } \right)$ .

Note : 1. Students must know the meaning of convergence and divergence of a series before proceeding with the solution.
2.The nth partial sum of the series for $n$ as $\left[ {1,\infty } \right)$ $\sum\limits_{n = 1}^\infty {{a_n}}$ is given by ${S_n} = {a_1} + {a_2} + .... + {a_n}$ .If the sequence of these partial sums \left\\{ {{S_n}} \right\\} converges to L, then the sum of the series also converges to L. Similarly, if \left\\{ {{S_n}} \right\\} diverges to L , then the sum of the series also diverges.