Find the limit $\lim_{n\to\infty}\frac{1}{\sqrt{n}}\int\limits_1^n\ln(1+\frac{1}{\sqrt{x}})dx.$ Note that $\ln(1+\frac{1}{\sqrt{x}})$ tends to $0$ at $x\rightarrow\infty$ Please check my solution: The function $\ln(1+\frac{1}{\sqrt{x}})$ increases and is bounded from above, since $0<\frac{1}{\sqrt{x}}<1$ and $\ln(1+x)<x$ at $x>0$. Therefore, using Stirling's integral formula, we have: $$\frac{1}{\sqrt{n}}\int\limits_1^n\ln(1+\frac{1}{\sqrt{x}})dx= \frac{1}{\sqrt{n}}\int\limits_1^n\ln(\frac{\sqrt{x}+1}{\sqrt{x}})dx$$$$=\frac{1}{\sqrt{n}}\int\limits_1^n\ln(\sqrt{x}+1)dx-\frac{1}{\sqrt{n}}\int\limits_1^n\ln(\sqrt{x})dx$$$$=\frac{2}{3\sqrt{n}}(\sqrt{n}+1)-\frac{2}{3\sqrt{n}}+\frac{1}{2\sqrt{n}}=\frac{2}{3}+\frac{1}{2n\sqrt{n}}.$$ Hence we get that the limit is ${\frac{2}{3}}$.
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4You can just use $$ \log \left( {1 + \frac{1}{{\sqrt x }}} \right) = \frac{1}{{\sqrt x }} + \mathcal{O}!\left( {\frac{1}{x}} \right). $$ How did you get the last equality? How did you use Stirling's formula? Btw, the correct answer is $2$. – Gary Mar 28 '23 at 06:52
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1Integration by part also works. Integral has a closed form: $$\frac1{\sqrt n}\int_1^n\ln(1+x^{-\frac12})dx=1+\left(\sqrt n-\frac1{\sqrt n}\right)\ln\left(1+\frac1{\sqrt n}\right)-\frac{\ln n}{2\sqrt n}-\frac1{\sqrt n}$$ – Svyatoslav Mar 28 '23 at 11:07
3 Answers
I don't think your solution is correct. Here's a different solution-
Since $x^{-3/2}$ is integrable on $(0,+\infty)$, we get \begin{align*} n^{-1/2} \int_1^n \ln\left(1+\frac{1}{\sqrt{x}} \right)\ \mathrm{d}x &= \frac{1}{\sqrt{n}} \int_1^n \frac{1}{\sqrt{x}} -\frac{1}{2x} +\mathcal{O}(x^{-3/2}) \ \mathrm{d}x\\ &= \frac{2\sqrt{n}-2-\ln n}{\sqrt{n}} +o(1)\\ &\to 2\end{align*}
Alternately, you can also note that the integral diverges, and then use L'Hospital's rule.
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Using L'Hospital's Rule is fine but you don't need to establish divergence of integral for that. +1 – Paramanand Singh Mar 29 '23 at 01:40
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@ParamanandSingh shouldn't the limit be of the form $\frac\infty\infty$ to apply L'Hospital? – Sayan Dutta Mar 29 '23 at 12:58
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1Well the rule applies to $\frac{\text{anything}} {\infty} $. No need to check behavior of numerator. – Paramanand Singh Mar 29 '23 at 13:14
Let
$$I_n=\frac{1}{\sqrt{n}}\int_1^n\ln\left(1+\frac{1}{\sqrt{x}}\right)~\mathrm{d}x.$$
Substituting $t=\sqrt{x}$ we get that
$$I_n=\frac{2}{\sqrt{n}}\int_1^\sqrt{n}t\ln\left(1+\frac{1}{t}\right)~\mathrm{d}t.$$
Now as
$$\lim_{t\to\infty}t\ln\left(1+\frac{1}{t}\right)=\lim_{h\downarrow0}\frac{\ln\left(1+h\right)}{h}=1,$$
it follows by the result in this post that
$$\lim_{n\to\infty}I_n=2.$$
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Under $nx\to x$, one has \begin{eqnarray} \frac{1}{\sqrt{n}}\int\limits_1^n\ln(1+\frac{1}{\sqrt{x}})dx=\sqrt n\int_{\frac1n}^1\ln(1+\frac{1}{\sqrt{nx}})dx. \end{eqnarray} Define $$ f_n(x)=\sqrt n\ln(1+\frac{1}{\sqrt{nx}})I_{[1/n,1]}(x) $$ Clearly $$ \lim f_n(x)=\frac{1}{\sqrt x}, |f_n(x)|\le \frac1{\sqrt x}$$ and $\frac1{\sqrt x}$ is integrable in $[0,1]$. By DCT, one has $$ \lim\frac{1}{\sqrt{n}}\int\limits_1^n\ln(1+\frac{1}{\sqrt{x}})dx=\lim\int_{0}^1f_n(x)dx=\int_0^1\lim f_n(x)dx=2.$$
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