Question

How do you test for convergence for $\sum {\dfrac{{3n + 7}}{{2{n^2} - n}}}$ for $n$ is $1$ to infinity?

Answer 1

Hint : As we know that convergence in mathematics, is a property of approaching a limit when the variable in a function increases or decreases. It is exhibited by some of the infinite functions and series. In this question, we will break the parts in two and then solve it. We should know that if the larger series converges, then the smaller series also converges, and if the smaller series diverges, then the larger series diverges.

Complete step-by-step answer :
In the above question we can write
$\sum\limits_{n = 1}^\infty {\dfrac{{3n + 7}}{{2{n^2} - n}}}$ .
We will break the fraction into two parts with same denominators: $\sum\limits_{n = 1}^\infty {\dfrac{{3n}}{{2{n^2} - n}} + \sum\limits_{n = 1}^\infty {\dfrac{7}{{2{n^2} - n}}} }$ .
Now the first part i.e. $\sum\limits_{n = 1}^\infty {\dfrac{{3n}}{{2{n^2} - n}}}$ , we will take the constant part out: $3\sum\limits_{n = 1}^\infty {\dfrac{1}{{2n - 1}}}$ ,
By putting the value of $n = 1$ we have: $3\sum\limits_{n = 1}^\infty {\dfrac{1}{{2 \times 1 - 1}}} = 3\sum\limits_{n = 1}^\infty {\dfrac{1}{n}}$ ,( we can write $\dfrac{1}{1}$ as $\dfrac{1}{n}$ because $n = 1$ .)
Now moving to the second part we have $\sum\limits_{n = 1}^\infty {\dfrac{7}{{2{n^2} - n}}}$ , similarly by taking the constant part out we have $7\sum\limits_{n = 1}^\infty {\dfrac{1}{{2{n^2} - n}}}$ .
We can write this series converge as $\int\limits_1^\infty {\dfrac{{dn}}{{2{n^2} - {1^{}}}}} = \ln (\left| {2n - 1} \right|) - \ln (n){]^\infty }_0 = \ln (2)$ .
Hence we can write it as diverge $+$ converge $=$ diverge.

Note : We should know that if the limit of a sequence goes to $\pm$ infinity, then the sequence is said to be divergent. But the sequence converges if the limit as $n$ approaches infinity i.e. constant numbers like $0,\pi$ . Before solving this type of question we should be fully aware of the convergent and divergent series.