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

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.