Question

How do you know if $ \sum {\dfrac{n}{{e\left( {{n^2}} \right)}}} $ converges from 1 to infinity?

Answer 1

Hint: In order to determine the correct option, find out the reciprocal of every given option , you will see that the for the options A and B the reciprocal exists and for the option C which is zero ,the reciprocal does not exists as there is no such number when multiplied with zero gives 1 and also $ \dfrac{1}{0} $ is undefined.

Complete step by step solution:
We are given a series in the summation form as $ \sum {\dfrac{n}{{e\left( {{n^2}} \right)}}} $ from 1 to infinity.
Since we have given the summation is from 1 to infinity , let's rewrite the summation with proper representation as
  $ = \sum\limits_{n = 1}^\infty {\dfrac{n}{{e\left( {{n^2}} \right)}}} $
As we know that we can pull out the constant from the summation to out of the summation. So pulling out the $ \dfrac{1}{e} $ from the summation ,we get
 $ = \dfrac{1}{e}\sum\limits_{n = 1}^\infty {\dfrac{n}{{{n^2}}}} $
Now using the property of exponent as when then base of numerator and denominator are same , then the exponent values get subtracted as $ \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}} $ . We get our expression as
\[
   = \dfrac{1}{e}\sum\limits_{n = 1}^\infty {{n^{1 - 2}}} \\
   = \dfrac{1}{e}\sum\limits_{n = 1}^\infty {{n^{ - 1}}} \\
   = \dfrac{1}{e}\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \;
 \]
Let’s expand the summation up to some terms to see the nature of the series
\[ = \dfrac{1}{e}\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + ........} \right)\]
Now trying to group the terms and replacing the terms in each group by the smaller term in the group, we get
\[
   = \dfrac{1}{e}\left( {1 + \dfrac{1}{2} + \left( {\dfrac{1}{3} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}} \right) + ........} \right) \\
   = \dfrac{1}{e}\left( {1 + \dfrac{1}{2} + \left( {\dfrac{1}{4} + \dfrac{1}{4}} \right) + \left( {\dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}} \right) + ........} \right) \\
   = \dfrac{1}{e}\left( {1 + \dfrac{1}{2} + \left( {\dfrac{1}{2}} \right) + \left( {\dfrac{1}{2}} \right) + ........} \right) \;
 \]
Since there are infinitely many $ \dfrac{1}{2}'s $ ,
\[
   = \dfrac{1}{e}\left( \infty \right) \\
   = \infty \;
 \]
From the above we can see that the summation given is a divergent harmonic series .
Therefore, the given $ \sum {\dfrac{n}{{e\left( {{n^2}} \right)}}} $ is a divergent harmonic series.

Note: 1. Expand the terms properly.
2. The replacement of the group with the smallest term in the group is to get the understanding of the nature of the series.
3. The value of exponential constant $ e $ is $ 2.71828 $ .