Question

How to find the sum of the harmonic sequence below? $\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{9}+\dfrac{1}{12}$

Answer 1

Here we have been given a harmonic sequence $\dfrac{1}{3},\dfrac{1}{6},\dfrac{1}{9},\dfrac{1}{12}$ . We need to find the sum of all the terms in the sequence. For doing that we can use the formula for finding the sum of terms in harmonic series given as $\dfrac{1}{d}\left( \ln \dfrac{\left( 2a+\left( 2n-1 \right)d \right)}{2a-d} \right)$ .

Complete step by step solution:
Now considering from the question we have been given a harmonic sequence $\dfrac{1}{3},\dfrac{1}{6},\dfrac{1}{9},\dfrac{1}{12}$ . We need to find the sum of all the terms in the sequence.
For doing that we can use the formula for finding the sum of the terms in a harmonic series mathematically given as $\dfrac{1}{d}\left( \ln \dfrac{\left( 2a+\left( 2n-1 \right)d \right)}{2a-d} \right)$ .
Here $a=3$, $n=4$ and $d=3$ .
Now by substituting these values we will have $\Rightarrow \dfrac{1}{3}\left( \ln \dfrac{\left( 6+\left( 7 \right)3 \right)}{3} \right)=\dfrac{1}{3}\ln 9\Rightarrow \dfrac{2.19}{3}=0.69$which is approximately equal to the value of $\dfrac{25}{36}$ .

Therefore we can conclude that the sum of all terms in the given harmonic sequence is $\dfrac{25}{36}$ .

Note: During the process of answering questions of this type we should be sure with our concepts that we apply and calculations that we perform. This is a very easy question that can be answered in a short span of time with very few mistakes. This can also be simply answered by simply adding all the four terms. The Least common multiple of $3,6,9,12$ is $36$ . The sum of the four terms will be given as $\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{9}+\dfrac{1}{12}=\dfrac{12+6+4+3}{36} = \dfrac{25}{36}$ . We have got the same answer in both the cases.