Natasha Henry

2022-01-25

Determine whether the given sequence is convergent or divergent
${a}_{n}=\frac{\mathrm{ln}\left(n+1\right)}{\sqrt{n}}$

Johnny Cummings

Expert

${a}_{n}=\frac{\mathrm{ln}\left(n+1\right)}{\sqrt{n}}$
Now,
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(n+1\right)}{\sqrt{n}}$, $\left[\frac{\mathrm{\infty }}{\mathrm{\infty }}\right]$ form
$=\underset{n\to \mathrm{\infty }}{lim}=\frac{\frac{1}{\left(n+1\right)}}{\frac{1}{2\sqrt{n}}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{2\sqrt{n}}{n+1}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{2\sqrt{n}}{n\left(1+\frac{1}{n}\right)}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{2\sqrt{n}}{\sqrt{n}\sqrt{n}\left(1+\frac{1}{n}\right)}$
$=\frac{2}{\mathrm{\infty }\left(1+\frac{1}{n}\right)}=0$
Thus, $\underset{n\to \mathrm{\infty }}{lim}\frac{\mathrm{ln}\left(n+1\right)}{\sqrt{n}}=0$
Hence, ${a}_{n}$ converhes to $0$

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