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Let's consider the function $f_{n}(x)$ with $x>0$ defined as:


I'd like to know what is the way to follow such that I may find out for what values of $x$ the function reaches its minimum and maximum , and then to compute these values. It's a
problem that came to my mind after studying another limit. However, I have more questions regarding this function, but for the moment it's enough if I find an answer for the part with minimum/maximum. Thanks.

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up vote 4 down vote accepted

First write \begin{align*} \log f_n(x) &= \sum_{j=1}^n \log(n^{x+1}+j^x) - \sum_{j=1}^n \log(n^{x+1}-j^x) \\ &= \sum_{j=1}^n \log\bigg( 1+\frac{j^x}{n^{x+1}} \bigg) + \sum_{j=1}^n \log\bigg( 1- \frac{j^x}{n^{x+1}} \bigg)^{-1}. \end{align*} The following inequalities are valid for all $0<y<\frac12$: \begin{align*} y-y^2 &< \log(1+y) < y \\ y &< \log(1-y)^{-1} < y+y^2 \end{align*} Therefore when $n\ge2$, \begin{align*} \log f_n(x) &> \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} - \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) + \sum_{j=1}^n \frac{j^x}{n^{x+1}} \\ &> 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \sum_{j=1}^n \frac1{n^2} \\ &= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \frac1n \end{align*} and

\begin{align*} \log f_n(x) &< \sum_{j=1}^n \frac{j^x}{n^{x+1}} + \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} + \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) \\ &< 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \sum_{j=1}^n \frac1{n^2} \\ &= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \frac1n. \end{align*} By the squeeze theorem, $$ \lim_{n\to\infty} \log f_n(x) = 2 \lim_{n\to\infty} \sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x = 2 \int_0^1 u^x\, du = \frac2{x+1}, $$ because the sum is a Riemann sum for the integral. In other words, $\lim_{n\to\infty} f_n(x) = e^{2/(x+1)}$ for all $x>0$ (since exponentiation is continuous).

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awesome. Thank you! (+1) – user 1618033 Aug 27 '12 at 7:19

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