Question

b) Find whether the following series are convergent or divergent √ 𝟏 /𝟒 + √ 𝟐 /𝟔 + √ 𝟑 /8 + ...

Answer 1

$\sqrt{\frac{1}{4}}+\sqrt{\frac{2}{6}}+\sqrt{\frac{3}{8}}+...=\sum_{n=1}^{\infty}\sqrt{\frac{n}{2(n+1)}}\\ a_n=\sqrt{\frac{n}{2(n+1)}}, a_{n+1}=\sqrt{\frac{n+1}{2(n+1+1)}}=\sqrt{\frac{n+1}{2(n+2)}}\\ \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_{n}}=\lim\limits_{n\to\infty}\frac{\sqrt{\frac{n+1}{2(n+2)}}}{\sqrt{\frac{n}{2(n+1)}}}=\\ =\lim\limits_{n\to\infty}\frac{n+1}{\sqrt{n(n+2)}}=1\\ \lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\sqrt{\frac{n}{2(n+1)}}=\\ =\lim\limits_{n\to\infty}\sqrt{\frac{n}{2n(1+\frac{1}{n})}}=\sqrt{\frac{1}{2}}\neq0$

The series are divergent.