b) Find whether the following series are convergent or divergent √ 𝟏 /𝟒 + √ 𝟐 /𝟔 + √ 𝟑 /8 + ...
14+26+38+...=∑n=1∞n2(n+1)an=n2(n+1),an+1=n+12(n+1+1)=n+12(n+2)limn→∞an+1an=limn→∞n+12(n+2)n2(n+1)==limn→∞n+1n(n+2)=1limn→∞an=limn→∞n2(n+1)==limn→∞n2n(1+1n)=12≠0\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}}\neq041+62+83+...=∑n=1∞2(n+1)nan=2(n+1)n,an+1=2(n+1+1)n+1=2(n+2)n+1n→∞limanan+1=n→∞lim2(n+1)n2(n+2)n+1==n→∞limn(n+2)n+1=1n→∞liman=n→∞lim2(n+1)n==n→∞lim2n(1+n1)n=21=0
The series are divergent.
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