Supremum and infimum of $\left\{(n^2+2n+1)^{\frac{1}{n^2}} \mid n \in\mathbb N \right\}$

Question

Task is to find infimum and supremum of $\left\{(n^2+2n+1)^{\frac{1}{n^2}} \mid n \in\mathbb N \right\}.$

I start from calculating derivative of $ f:\mathbb{R} \rightarrow \mathbb{R}$ where $ f(n) = (n^2+2n+1)^{\frac{1}{n^2}}$to find $n_{0} \in \mathbb{R}$ where derivative is equal to $0$.

\begin{align} ((n^2+2n+1)^\frac{1}{n^2})' &= \left(e^{\ln\left((n+1)^\frac{2}{n^2}\right)}\right)' \\ &= \left(e^{\frac{2}{n^2}\cdot\ln(n+1)}\right)' \\ &= (n+1)^{\frac{2}{n^2}}\cdot \bigg(\frac{2}{n^2}\cdot\ln(n+1)\bigg)' \\ &= (n+1)^{\frac{2}{n^2}}\cdot\bigg(\bigg(\frac{2}{n^2}\cdot(\ln(n+1))'\bigg)+\bigg(\bigg(\frac{2}{n^2}\bigg)'\cdot\ln(n+1)\bigg) \\ &= (n+1)^{\frac{2}{n^2}}\cdot\bigg(\bigg(\frac{2}{n^2}\cdot\frac{1}{n+1}\bigg)+\bigg(\frac{-4}{n^3}\cdot\ln(n+1)\bigg) \\ &= (n+1)^{\frac{2}{n^2}}\cdot\bigg(\frac{2n + -4(n+1)\cdot\ln(n+1)}{n^3\cdot(n+1)}\bigg) \end{align}

Where derivative is equal $0$ there may be extremum

$$(n+1)^{\frac{2}{n^2}}\cdot\frac{2n + -4(n+1)\cdot\ln(n+1)}{n^3\cdot(n+1)} = 0 \\ \Leftrightarrow 2n + -4(n+1)\cdot\ln(n+1) = 0$$

I think that supremum is maximum of $\lim_{n\rightarrow \infty} (n+1)^{\frac{2}{n^2}} = 1$, value in $1$ ($=4$) and values of $n\in \mathbb{N} $ nearest $n_{0}$.

How to solve this equation? Do I think in the right direction?

Answer 1

If you look carefully at what you've done, it implies that the derivative is actually negative as long as $n \geq 1$ (which is the range you care about). This means the supremum is the value at $n=1$ and the infimum is the limit at $\infty$.

Answer 2

Computation-free way to see the expression is decreasing: The expression equals $[(n+1)^2]^{1/n^2}.$ Apply $\ln$ and then replace $n$ by $x$ (so we can think of of a continuous variable). We get $[2\ln(x+1)]/x^2.$ Now

$$\frac{\ln (x+1)}{x} = \frac{1}{x}\int_1^{1+x}\frac{dt}{t}.$$

This is the average of a decreasing function over the interval $[1,1+x].$ That has to be a decreasing function of $x,$ because as $x$ gets larger, smaller values of $1/t$ are included in the mix. We also have $1/x$ decreasing. The product of positive decreasing functions is decreasing. So the $\ln$ of our expression is decreasing, hence the expression itself is decreasing.