limit of $u_{n} =\int_{0}^{1} \frac{1}{1+t+t^2+...+t^n} \, \mathrm{d}t$ find the limit of
$$u_{n} =\int_{0}^{1} \frac{1}{1+t+t^2+...+t^n} \, \mathrm{d}t$$
I recognise the inverse of the sum of a geometric serie (with t as the common ratio), and with a continuous extension in 1
$$u_{n} =\int_{0}^{1} \frac{t-1}{t^{n+1} -1} \, \mathrm{d}t$$ and I cannot go further.
 A: For $t\in[0,1[,\lim\limits_{n\rightarrow +\infty}\frac{1}{1+t+\ldots+t^n}=1-t$, and
$\frac{1}{1+t+\ldots+t^n}\leqslant 1$ for all $t\in[0,1]$ and $n\in\mathbb{N}$ thus, using the dominated convergence theorem, we have
$$ \lim\limits_{n\rightarrow +\infty}\int_0^1\frac{dt}{1+t+\ldots+t^n}=\int_{[0,1[}(1-t)dt=\frac{1}{2} $$
A: First, it's clear that
$$
\frac{1}{1+t+t^2 +\dots+t^n} \leq 1
$$
for all $n$ and all $t \in [0,1]$.  Therefore you can use the Dominated Convergence Theorem (or just the special case which is sometimes called the Bounded Convergence Theorem) to interchange the limit and the integral:
$$
\lim_{n \to \infty} \int_0^1 \frac{t-1}{t^{n+1}-1} \,dt = \int_0^1 \lim_{n \to \infty} \frac{t-1}{t^{n+1}-1} \,dt = \int_0^1 (1-t) \,dt = \frac12.
$$
A: No need to invoke Dominated convergence theorem. We estimate
$$
\frac{1}{n+2}=\int_0^1 \frac{1-t}{1-t}t^{n+1}\geq \int_0^1 \frac{1-t}{1-t^{n+1}}t^{n+1}=u_n-\int_0^1 1-t \,dt = u_n-\frac{1}{2}\geq 0.
$$
Now we send $n\to +\infty$.
A: For what it's worth, we may avoid using Dominated Convergence Theorem. Let $f_n(t)=\frac{1}{1+t+t^2+\cdots+t^n}-(1-t)$ and $v_n=\int_0^1f_n(t)dt$. Then $\{v_n\}$ converges because it is a decreasing sequence that is bounded below. Now, for any $a\in(0,1]$, we have
$$0\le v_n\le2a+\int_a^1f_n(x)dx.\tag{1}$$
Note that $\{f_n\}$ is a monotonic decreasing sequence of continuous real-valued function that converges pointwise to $0$ on the compact set $[a,1]$. The convergence is therefore uniform, by Dini's Theorem. So, if we pass $n$ to the limit in $(1)$, we can interchange the order of limit and integration on the RHS. Hence $0\le \lim_{n\to\infty}v_n\le 2a$ for every $a\in(0,1]$, i.e. $\lim_{n\to\infty}v_n=0$. It follows that $\lim_{n\to\infty}u_n=\int_0^1(1-t)dt=\frac12$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

$\ds{\large\left.a\right)\ A\ short\ one:}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\,\int_{0}^{1}{\dd t \over 1 + t + t^{2} + \cdots + t^{n}}} =
\lim_{n \to \infty}\,\int_{0}^{1}{1 - t \over 1 - t^{n + 1}}\,\dd t
\\[5mm] = &\
\underbrace{\int_{0}^{1}\pars{1 - t}\dd t}_{\ds{1 \over 2}}\ +\
\underbrace{\lim_{n \to \infty}\,\int_{0}^{1}\pars{1 - t}{t^{n + 1} \over 1 - t^{n + 1}}\,\dd t}_{\ds{0}}
\\[5mm] = &\ \bbx{\large{1 \over 2}} \\ &\
\end{align}
Note that
\begin{align}
0 & <
\verts{\int_{0}^{1}\pars{1 - t}{t^{n + 1} \over
1 - t^{n + 1}}\,\dd t} <
\int_{0}^{1}\pars{1 - t}{t^{n + 1} \over 1 - t}\,\dd t
\\[5mm] & = {1 \over n + 2}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,
{\large\color{red}{0}}
\end{align}

$\ds{\large\left.b\right)\ An\ alternative:}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\,\int_{0}^{1}{\dd t \over 1 + t + t^{2} + \cdots + t^{n}}} =
\lim_{n \to \infty}\,\int_{0}^{1}{1 - t \over 1 - t^{n + 1}}\,\dd t
\\[5mm] \stackrel{t^{n + 1}\,\,\,\, \mapsto\ t}{=}\,\,\, &\
\lim_{n \to \infty}\,\int_{0}^{1}{1 - t^{1/\pars{n + 1}} \over 1 - t}
\bracks{{t^{1/\pars{n + 1} - 1\,\,} \over n + 1}}\dd t
\\[5mm] = &
\lim_{n \to \infty}\left\{{1 \over n + 1}%
\left[%
\int_{0}^{1}{1 - t^{\pars{1 - n}/\pars{1 + n}} \over 1 - t}
\,\,\dd t\right.\right.
\\[2mm] &\ \phantom{\lim_{n \to \infty}{1 \over n + 1}\,\,\,}\left.\left.-
\int_{0}^{1}{1 - t^{-n/\pars{n + 1}} \over 1 - t}
\,\,\dd t\right]\right\}
\\[5mm] = &\
\lim_{n \to \infty}\braces{{1 \over n + 1}%
\left[\Psi\pars{2 \over n + 1}\right.
\left.- \Psi\pars{1 \over n + 1}\right]}
\label{1}\tag{1}
\end{align}
where I used A & S table identity $\color{black}{\bf 6.3.22}$. $\ds{\Psi}$ is the
Digamma Function.
With the $\ds{\Psi}$-recurrence formula
( see $\ds{\color{black}{\bf 6.3.5}}$ in A & S table ):
$$
\left\{\begin{array}{rclcl}
\ds{\Psi\pars{2 \over n + 1}} & \ds{=} &
\ds{\Psi\pars{n + 3 \over n + 1} - {n + 1 \over 2}}
& \ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}} &
\ds{\Psi\pars{1} - {n \over 2}}
\\[2mm]
\ds{\Psi\pars{1 \over n + 1}} & \ds{=} &
\ds{\Psi\pars{n + 2 \over n + 1} - \pars{n + 1}}
& \ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}} &
\ds{\Psi\pars{1} - n}
\end{array}\right.
$$
(\ref{1}) becomes
\begin{align}
& \mbox{} \\
&\bbox[5px,#ffd]{\lim_{n \to \infty}\,\int_{0}^{1}{\dd t \over 1 + t + t^{2} + \cdots + t^{n}}} =
\bbx{\large{1 \over 2}} \\ &
\end{align}
