Question

Derive the formula for the sum of \[{1^{st}}\]\[n\] natural numbers in harmonic progression.

Answer 1

Hint: In this problem, we have to derive the formula for the first \[n\] natural number in harmonic progression H.P. We know that the arithmetic sequence which increases with the constant value means arithmetic sequence. And then, the harmonic sequence is formed by taking reciprocals of an arithmetic progression.
Formulae used in this problem :
\[{S_n} = \left( {{a_1} + \dfrac{{\left( {n - 1} \right)d}}{2}} \right) \times n\];
\[d = {a_2} - {a_1} = {a_3} - {a_2}\];
Where
\[{a_1} = \] the first term
 \[d = \] common difference.

Complete step-by-step solution:
In the given problem we have to derive the formula for the first \[n\] natural number in harmonic progression.
We know that the harmonic progression is the reciprocals of an arithmetic progression.
In the arithmetic sequence, it increases with a constant value and has a common difference.
In the question, the data is given, natural numbers
Natural numbers start from one, two, three and so on…
The sequence of the first \[n\] natural numbers is \[1,2,3..............\left( {n - 1} \right),n\]
We know that the \[{a_1} = 1\]
And we have to find the common difference. We know the formula for finding common differences.
The common difference that is equal to the first term is subtracting from the second term.
Now apply these values to the formula, we get
\[d = 2 - 1 = 3 - 2\]
Solve this we get,
\[d = 1\]
We know that the formula,
\[{S_n} = \left( {{a_1} + \dfrac{{\left( {n - 1} \right)d}}{2}} \right) \times n\]
Now apply the values,
\[ = \left( {1 + \dfrac{{\left( {n - 1} \right)1}}{2}} \right) \times n\]
Now take the least common multiple(L.C.M) for the terms we get,
\[ = \left( {\dfrac{{2 + \left( {n - 1} \right)}}{2}} \right)n\]
Now multiply the terms which are inside the bracket by \[n\], we get
\[=\left( {\dfrac{{2n + {n^2} - n}}{2}} \right)\]
Now simplify this we get,
\[ = \left( {\dfrac{{{n^2} + n}}{2}} \right)\]
Now take common terms out
\[ = n\left( {\dfrac{{n+1}}{2}} \right)\]
Therefore
\[{S_n} = n\left( {\dfrac{{n+1}}{2}} \right)\]
This is for the arithmetic progression. But we have to find this for harmonic progress.
Already we know that the harmonic progression takes the reciprocal from the arithmetical progression.
Therefore the formula for the harmonic progression is \[ = \dfrac{2}{{n\left( {n+1} \right)}}\]
Hence the formula derived.

Note: In mathematics, a harmonic progression is a progression formed by taking the reciprocal of an arithmetic progression. An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is a constant value. A series of numbers in which each number is multiplied or divided by the constant number to produce the next number in the sequence.