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$\lim\limits_{n\rightarrow\infty}\sum_{k = 1}^{n} \frac{1}{(k+n)\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}$. I can find it using integral: $\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{k = 1}^{n} \frac{1}{(1 + \frac{k}{n})\sqrt{1 + n\ln({1+\frac{k}{n^2}})}}$, but how get rid of $n$ and $\frac{k}{n^2}$ in sqrt? Gan you give me a hint, please?

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We have \begin{align} \frac 1{\sqrt{1 + n\ln\left({1+\frac{k}{n^2}}\right)}} &=\left(1 + n\ln\left({1+\frac{k}{n^2}}\right)\right)^{-1/2}\\ &=\left(1 + \frac kn+O\left(\frac{k^2}{n^3}\right)\right)^{-1/2}\\ &=\frac 1{\sqrt{1 + \frac kn}}+\frac 1nO\left(\frac{k}{n}\right)^2 \end{align} Consequently, \begin{align} \sum_{k = 1}^{n} \frac{1}{(k+n)\sqrt{1 + n\ln({1+\frac{k}{n^2}})}} &=\frac 1n\sum_{k = 1}^{n} \frac{1}{(1+\frac kn)\sqrt{1 + \frac kn}}+\frac 1n\sum_{k = 1}^{n} \frac{O\left(\frac{k^2}{n^3}\right)}{(1+\frac kn)\sqrt{1 + \frac kn}}\\ &=\frac 1n\sum_{k = 1}^{n}\left(1+\frac kn\right)^{-3/2}+\frac 1{n^2}\sum_{k = 1}^{n} O\left(\frac{k^2}{n^2}\right)\left(1+\frac kn\right)^{-3/2} \end{align} and the last sum vanish as $n\to\infty$.

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Hint

When $n$ is large $$\log \left(1+\frac{k}{n^2}\right)\sim \frac k {n^2}$$ this would make things nice, I hope !

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  • $\begingroup$ Bit it means that $n\log(1+\frac{k}{n^2}) = \frac{k}{n} + o(n)$ if I am not mistaken. So, I need to get rid of o(n) $\endgroup$ – ErlGrey Mar 19 at 10:47
  • $\begingroup$ Just as @Fabio Lucchini answered. $\endgroup$ – Claude Leibovici Mar 19 at 11:17
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By definition of Riemann integral we have $$\int_{0}^{1}f(x)\,dx=\lim_{n\to \infty} \frac {1}{n}\sum_{k=1}^{n}f(t_k),\,\frac{k-1}{n}\leq t_k\leq\frac{k} {n} \tag{1}$$ provided the function $f:[0, 1]\to \mathbb {R} $ is Riemann integrable on $[0,1]$.

Further as a result of Duhamel Principle if functions $f:[0, 1]\to\mathbb {R}, g:[0, 1]\to\mathbb {R} $ are Riemann integrable on $[0,1]$ then $$\int_{0}^{1}f(x)g(x)\,dx=\lim_{n\to \infty} \frac {1}{n}\sum_{k=1}^{n}f(t_k)g(t'_k), \, \frac{k-1}{n}\leq t_k, t'_k\leq \frac{k} {n} \tag{2}$$ Now we observe that $$\frac{k/n} {1+k/n^2}\leq n\log\left(1+\frac{k}{n^2}\right)\leq \frac{k} {n} $$ and clearly $$\frac{k-1}{n}\leq\frac{k/n}{1+k/n^2}=\frac{kn}{n^2+k}$$ We can now take $$f(x) =\frac{1}{1+x},g(x)=\frac{1}{\sqrt{1+x}},t_k=\frac{k}{n},t'_k=n\log\left(1+\frac{k}{n^2}\right)$$ and the expression under limit in question equals $$\frac{1}{n}\sum_{k=1}^{n}f(t_k)g(t'_k)$$ which by $(2)$ tends to $$\int_{0}^{1}f(x)g(x)\,dx=\int_{0}^{1}(1+x)^{-3/2}\,dx=2-\sqrt{2}$$

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