Approximate value of \({\left( {1 + \frac{1}{n}} \right)^n}\)

If we actually calculate the values \({\left( {1 + \frac{1}{{100}}} \right)^{100}}\) or \({\left( {1 + \frac{1}{{1000}}} \right)^{1000}}\) with the help of calculator or log table, we will come to the conclusion that these values will be very close to 2.7. Actually  the value of \({\left( {1 + \frac{1}{n}} \right)^n}\) is very close to \(e\), where \(e = 2.7182\) and \(n\) is a large number.

For example  \({\left( {1 + \frac{1}{{1000}}} \right)^{1000}}\)= \({\left( {1 + \frac{1}{{500}}} \right)^{500}} = 2.71\).

\({\left( {1 + \frac{1}{n}} \right)^n} = e\)  

and  \({\left( {1 - \frac{1}{n}} \right)^n} = {e^{ - 1}} = \frac{1}{e}\)

where \(n\) is a large number.

Using this result, we can solve some questions very quickly, for example suppose we want to calculate the approximate value of the number \({(0.99)^{100}}\), write the number as \({(1 - 0.01)^{100}} = {\left( {1 - \frac{1}{{100}}} \right)^{100}} = \frac{1}{e} = 0.36\), (putting the approximate value of \(e = 2.71\)).

Thus \({0.99^{100}} \approx 0.36\)

Example  01: If \(A = {400^{101}}\) and \(B = {404^{100}}\), find the approximate value of \(\frac{A}{B}\)

Maximum value of the number \({n^{1/n}}\)

If \(n\) is a positive number, then the value of \({n^{1/n}}\) is maximum when \(n = e \approx 2.7\), after that it decreases. If we consider only natural values for \(n\), then the maximum value of \({n^{1/n}}\) is obtained at \(n = 3\) as \(3\) is the closest number to \(e\). So among the numbers \({2^{1/2}},\,{3^{1/3}},\,{4^{1/4}},\,{5^{1/5}},\,\,{6^{1/6}},\,{7^{1/7}}\), the number \({3^{1/3}}\) is the largest. After reaching to the maximum value, the number \({n^{1/n}}\) starts decreasing. At bigger values of \(n\), we get a smaller value of \({n^{1/n}}\). So \({7^{1/7}} < {6^{1/6}} < {5^{1/5}}\).

Example 02: Arrange them in descending order \(a = {2^{1/2}},\,\,b = {3^{1/3}},\,\,c = \,{6^{1/6}},\,\,d = {9^{1/9}}\)