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Question Number 52682 by maxmathsup by imad last updated on 11/Jan/19

$$findnatureoftheserie\sum _{n=1}^{\mathrm{\infty }}\frac{\sqrt{n+1}-\sqrt{n}}{nln(n+1)}$$

Commented by Abdo msup. last updated on 12/Jan/19

$$let{u}_n=\frac{\sqrt{n+1}-\sqrt{n}}{nln(n+1)}wehave$$$$u_n=\frac{1}{(\sqrt{n+1}+\sqrt{n})nln(n+1)}$$$$\sqrt{n+1}+\sqrt{n}\sim 2\sqrt{n}andln(n+1)=ln\left(n(1+\frac{1}{n})\right)$$$$ln\left(n\right)+ln(1+\frac{1}{n})\sim ln\left(n\right)+\frac{1}{n}\Rightarrow$$$$u_n\sim \frac{1}{2\sqrt{n}(ln\left(n\right)+\frac{1}{n})}=\frac{1}{2\sqrt{n}ln\left(n\right)+\frac{2}{\sqrt{n}}}\sim \frac{1}{2\sqrt{n}ln\left(n\right)}$$$$thesequence{v}_n=\frac{1}{2\sqrt{n}ln\left(n\right)}isdecreasing\Rightarrow$$$$\mathrm{\Sigma }{v}_{n}and{\int }_2^{+\mathrm{\infty }}\frac{dx}{2\sqrt{x}ln\left(x\right)}havethesamenatureof$$$$convergencebut$$$${\int }_2^{+\mathrm{\infty }}\frac{dx}{2\sqrt{x}ln\left(x\right)}{=}_{ln\left(x\right)=t}{\int }_{ln\left(2\right)}^{+\mathrm{\infty }}\frac{e^{t}dt}{2e^{\frac{t}{2}}t}$$$$=\frac{1}{2}{\int }_{ln\left(2\right)}^{+\mathrm{\infty }}\frac{e^{\frac{t}{2}}}{t}dtandthisintegraldiverges$$$$({lim}_{t\to +\mathrm{\infty }}t.\frac{e^{\frac{t}{2}}}{t}=+\mathrm{\infty }+so\mathrm{\Sigma }{u}_{n}diverges.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jan/19

$$\frac{T_n}{T_{n-1}}=\frac{\sqrt{n+1}-\sqrt{n}}{nln(n+1)}\times \frac{(n-1)ln\left(n\right)}{\sqrt{n}-\sqrt{n-1}}$$$$$$$$=\frac{1-\sqrt{\frac{n}{n+1}}}{\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n-1}{n+1}}}\times (1-\frac{1}{n})\times \frac{lnn}{ln(n+1)}$$$$=\frac{1-\sqrt{\frac{1}{1+\frac{1}{n}}}}{\sqrt{\frac{1}{1+\frac{1}{n}}}}\times (1-\frac{1}{n})\times \frac{lnn}{ln(n+1)}$$$$whenn\to \mathrm{\infty }$$$$=\frac{1-\sqrt{\frac{1}{1+0}}}{\sqrt{\frac{1}{1+0}}}\times (1-0)\times \frac{lnn}{ln(n+1)}$$$$=0soconvergent...$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\underset{n\to \mathrm{\infty }}{\mathrm{li}p}\frac{T_n}{T_{n-1}}$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Jan/19

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