Question 154
Let $d_n = \sum_{k=0}^{2n} \frac{(-1)^k}{n+k} = \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \dots + \frac{1}{3n}$ for $n \ge 1$.
Prove that
$$
\frac{2}{3n+1} \lt d_n \lt \frac{2}{3n-2}.
$$
(Otgonbayar Uuye)
Prove that
$$
\frac{2}{3n+1} \lt d_n \lt \frac{2}{3n-2}.
$$
(Otgonbayar Uuye)