Question #311478

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


1
Expert's answer
2022-03-15T10:20:34-0400

14+26+38+...=n=1n2(n+1)an=n2(n+1),an+1=n+12(n+1+1)=n+12(n+2)limnan+1an=limnn+12(n+2)n2(n+1)==limnn+1n(n+2)=1limnan=limnn2(n+1)==limnn2n(1+1n)=120\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.


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