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If $\lim\limits_{n \to \infty}\frac{1}{n}\sum\limits_{k=1}^n k\ln \left(\frac{n^2+(k-1)^2}{n^2+k^2}\right)$ exists and equal to $l$, find $l$.

I don't know how to approach this probelm. I thought about sandwich theorem and Integration as limit of sum but failed becuase as $n$ approaches $\infty$ quantity written in logarithm tends to $1$ which make $\ln \left(\frac{n^2+(k-1)^2}{n^2+k^2}\right)=0$

Please Help me solve this problem

pie
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mathophile
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4 Answers4

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Put $L_n = \sum_{k=1}^n \frac{k}{n} \log\Big(\frac{n^2 + (k-1)^2}{n^2 + k^2}\Big)$, for the $n$th term, and define $f(x) = \log(1+x^2)$. Then, \begin{align*} L_n &= \sum_{k=1}^n \frac{k}{n} \log\Big(\frac{1 + (\tfrac{k-1}{n})^2}{1 + (\tfrac{k}{n})^2}\Big)\\ &= \sum_{k=0}^{n-1} \frac{k+1}{n} f\Big(\frac{k}{n}\Big) - \sum_{k=1}^n \frac{k}{n} f\Big(\frac{k}{n}\Big)\\ &= \sum_{k=1}^{n} \frac{1}{n} f\Big(\frac{k}{n}\Big) - \frac{n+1}{n} f(1). \end{align*} The first term on the righthand side above a Riemann summation. Thus, your limit is, $$ L = \lim_{n \to \infty} L_n = \int_0^1 \log(1+x^2) \, dx - \log(2) = \frac{\pi}{2} - 2. $$

Drew Brady
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We can note that if $f(x) =\log(1+x^2)$ then $f'(x) =2x/(1+x^2)$ and hence by mean value theorem $$\log\frac{1+(k-1)^2/n^2}{1+k^2/n^2}=-\frac{1}{n}\cdot\frac{2t_k}{1+t_k^2}$$ for some $t_k\in((k-1)/n,k/n)$.

The expression under limit can be then written as $$(-2)\cdot \frac{1}{n}\sum_{k=1}^n\frac{k} {n} \cdot\frac{t_k} {1+t_k^2}$$ Leaving the first factor $(-2)$ the rest of the expression is a (not so typical) Riemann sum for the function $$x\cdot\frac{x} {1+x^2}$$ over interval $[0,1]$ with uniform partition into $n$ subintervals.

The desired limit is then $-2\int_0^1\frac{x^2}{1+x^2}\, dx$ which can be easily calculated.

Paramanand Singh
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Let $f(x) = -\log(1+x^2)$. Then the sum inside the limit can be recast as:

$$ \sum_{k=1}^{n} \frac{k}{n}\left[ f\left(\frac{k}{n}\right) - f\left(\frac{k-1}{n}\right) \right] \to \int_{0}^{1} x \, \mathrm{d} g(x) = \int_{0}^{1} x g'(x) \, \mathrm{d}x $$

as $n \to \infty$

Sangchul Lee
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$$\ln(1+x) = \sum_{m=1}^ \infty \frac{(-1)^{m+1} x^m}{ m} \ \ \ \ \text{ with radius of convergence = 1 }$$

$$\lim\limits_{n\to\infty}\frac{1}{n} \sum\limits_{k=1}^n k\ln \left(\frac{n^2+(k-1)^2}{n^2+k^2}\right) = \lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n k \ln \left( 1 + \frac{1-2k}{n^2+k^2}\right)$$ $$= \lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n k \sum_{m=1}^ \infty \frac{(-1)^{m+1} \left( \frac{1-2k}{n^2+k^2}\right)^m}{ m} $$

$$\left| \frac{1}{n} \sum\limits_{k=1}^n k \sum_{m=1}^ \infty \frac{(-1)^{m+1} \left( \frac{1-2k}{n^2+k^2}\right)^m}{ m} - \frac{1}{n} \sum\limits_{k=1}^n k \left( 1 + \frac{1-2k}{n^2+k^2}\right)\right|$$

$$= \left|\frac{1}{n} \sum\limits_{k=1}^n k \sum_{m=2}^ \infty \frac{(-1)^{m+1} \left( \frac{1-2k}{n^2+k^2}\right)^m}{ m}\right|< \sum\limits_{k=1}^n \sum_{m=2}^ \infty {\left( \frac{2n+1}{n^2}\right)^m} $$

$$= n\left(\frac{2n+1}{n^2}\right)^2\cdot \frac{1}{1-\frac{2n+1}{n^2}} \ \ \ \ \ \ \text{Which goes to $0$ as $n $ goes to $\infty$}$$

By squeeze theorem $\displaystyle \lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n k \ln \left( 1 + \frac{1-2k}{n^2+k^2}\right)=\lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n k \left( \frac{1-2k}{n^2+k^2}\right) =\lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n k \left( \frac{-2k}{n^2+k^2}\right) $

$$\lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n k \left( \frac{-2k}{n^2+k^2}\right) = -2\lim\limits_{n \to \infty }\frac{1}{n} \sum\limits_{k=1}^n \frac{\frac{k^2}{n^2}}{1+\frac{k^2}{n^2}}$$

$$=-2\int_0^1 \frac{x^2}{1+x^2}dx= \frac{\pi}{2}-2$$

pie
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