Let a(n) be the integral from x = 1/2 to x = 2 of 2^x/(1 + x^2)^n dx. Prove that a(n+1)/a(n) tends to 4/5 as n tends to infinity.

Step-by-step solution

1. 1

   Rewrite the ratio as a weighted average. Put $g(x)=\dfrac{1}{1+x^{2}}$, so that $a_n=\int_{1/2}^{2}2^{x}g(x)^{n}\,dx$ and $a_{n+1}=\int_{1/2}^{2}g(x)\cdot 2^{x}g(x)^{n}\,dx$. Therefore $\frac{a_{n+1}}{a_n}=\frac{\int_{1/2}^{2}g(x)\,w_n(x)\,dx}{\int_{1/2}^{2}w_n(x)\,dx},\qquad w_n(x)=2^{x}g(x)^{n}>0,$ which is exactly the average of $g$ against the positive weight $w_n$. An average always lies between the minimum and the maximum of $g$, so the whole proof is about showing the weight concentrates where $g$ is largest.

2. 2

   Locate the maximum of the base function. On $x>0$ the function $g(x)=1/(1+x^{2})$ is strictly decreasing, so on $\left[\frac12,2\right]$ it attains its maximum at the left endpoint: $M=g\!\left(\tfrac12\right)=\frac{1}{1+\frac14}=\frac45 .$ This already identifies $4/5$ as the only candidate for the limit.

3. 3

   Get the upper bound for free. Since $g(x)\le M$ everywhere on the interval, $a_{n+1}=\int_{1/2}^{2}g(x)\,w_n(x)\,dx\le M\int_{1/2}^{2}w_n(x)\,dx=M\,a_n,$ so $\frac{a_{n+1}}{a_n}\le \frac45$ for every $n$, and hence $\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}\le\frac45$.

4. 4

   Show the mass concentrates at the left endpoint. Fix $\varepsilon>0$. By continuity choose $\delta>0$ with $g(x)>M-\varepsilon$ on $\left[\frac12,\frac12+\delta\right]$ and with $q:=g\!\left(\frac12+\delta\right)<M-\varepsilon$. Splitting $a_n=I_n+J_n$ at $\frac12+\delta$ and using monotonicity of $g$, $I_n\ge C_1 (M-\varepsilon)^{n},\qquad J_n\le C_2 q^{n},\qquad C_1=\int_{1/2}^{1/2+\delta}2^{x}dx,\ C_2=\int_{1/2+\delta}^{2}2^{x}dx .$ Since $q<M-\varepsilon$, the quotient $J_n/I_n\le \frac{C_2}{C_1}\left(\frac{q}{M-\varepsilon}\right)^{n}\to 0$, so the tail piece is negligible.

5. 5

   Conclude with the lower bound. Because almost all of the weight sits on $\left[\frac12,\frac12+\delta\right]$, where $g>M-\varepsilon$, the weighted average satisfies $\frac{a_{n+1}}{a_n}>M-2\varepsilon$ for all large $n$, so $\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}\ge M-2\varepsilon$. As $\varepsilon>0$ was arbitrary, $\liminf\ge\frac45$, and combined with the upper bound the limit equals $\frac45$.

6. 6

   Check the rate numerically. Convergence is genuinely slow, since the concentration width shrinks only like $1/n$: numerical quadrature gives $a_{21}/a_{20}\approx 0.762$, $a_{41}/a_{40}\approx 0.781$ and $a_{61}/a_{60}\approx 0.787$, all below $0.8$ and increasing toward it, exactly as the two bounds above predict.