Today, I was practicing some question then this question came in my mind.
How can we find the sum of first n terms of the series which is of the form,
$\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{5}{6}+…\sf upto\, nth\, term$
Just like we have formulas,
$\sum\limits_{i=1}^ni = \dfrac{n(n+1)}{2}$
$\sum\limits_{i=1}^ni^2 = \dfrac{n(n+1)(2n+1)}{6}$
Any hint would be helpful to me.
Best Answer
You may write the series as $$\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{6}\right)+\ldots+\left(1-\frac{1}{2n}\right)=n-\frac{1}{2}H_n$$
I'm not aware of any nice form that doesn't use integrals.