Compute the $\lim_{n\to\infty}\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}$ I know you should prove $f_n (x) = \frac1{1+x+\frac{x^n}n}$ converges uniformly and then the limit follows immediately however I’m struggling to prove the uniform convergence. 
 A: In this case you can even compute it directly:
$\displaystyle 0\le \int_0^1 \frac{1}{1+x}dx-\int_0^1 \frac{1}{1+x+\frac{x^n}{n}}dx\\\displaystyle=\int_0^1 \frac{\overbrace{x^n}^{\le 1}}{\underbrace{(n+nx+x^n)}_{\ge n}\underbrace{(1+x)}_{\ge 1}}dx\le \int_0^1 \frac{1}{n}dx=\frac 1n\to 0$
A: It suffices to show $x^n/n$ converges uniformly to $0$, since $1+x>0$ on $[0,1]$. For all $x\in[0,1]$, $x\leq 1$, implying $x^n\leq 1$ as well...
Alternatively, since $\frac{1}{1+x+x^n/n}\leq \frac{1}{1+x}$ and the latter is integrable on $[0,1]$, you could use dominated convergence.
A: $\begin{array}\\
f_n
&=\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}\\
&\lt\int_0^1\frac{\mathrm{d}x}{1+x}\\
&= \ln(2)\\
\end{array}
$
and
$\begin{array}\\
f_n
&=\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}\\
&\gt\int_0^1\frac{\mathrm{d}x}{1+x+\frac1{n}}\\
&=\ln(2+\frac1{n})-\ln(1+\frac1{n})\\
&=\ln(2)+\ln(1+\frac1{2n})-\ln(1+\frac1{n})\\
&=\ln(2)+\ln(\frac{2n+1}{2n})-\ln(\frac{n+1}{n})\\
&=\ln(2)+\ln(\frac{2n+1}{2(n+1)})\\
&=\ln(2)+\ln(1-\frac1{2(n+1)})\\
&>\ln(2)-\dfrac{\frac1{2(n+1)}}{1-\frac1{2(n+1)}}
\qquad\text{since } \ln(1-x) > -\dfrac{x}{1-x} (*)\\
&=\ln(2)-\dfrac{1}{2(n+1)-1}\\
&=\ln(2)-\dfrac{1}{2n+1}\\
\end{array}
$
so
$f_n \to \ln(2)$.
$(*)\ -\ln(1-x)
=\int_{1-x}^1 \dfrac{dx}{x}
\lt \dfrac{x}{1-x}
$
