Evaluate $ \lim\limits_{n \to \infty}\sum\limits_{k=2}^{n} \frac{1}{\sqrt[k]{n^k+n+1}+1} $ $$ \lim_{n \to \infty}\sum_{k=2}^{n} \frac{1}{\sqrt[k]{n^k+n+1}+1} $$
I expect the squeeze theorem helps us solving this but I can't find the inequality.
The result should be $1$.

Consider
$s(n)
=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}
$
where
$f(n) \ge 0$ and
$f(n)/n^{c} \to 0$
for some $c > 0$.
$\begin{array}\\
s(n)
&=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}\\
&\lt\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k}}\\
&=\sum_{k=2}^{n} \dfrac{1}{n}\\
&\to \ln(n)-1+\gamma\\
\end{array}
$
Similarly,
since
$(1+x)^k \ge 1+kx,
(1+x/k)^k \ge 1+x$
so
$(1+x)^{1/k} \le 1+x/k$,
$\begin{array}\\
s(n)
&=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}\\
&=\sum_{k=2}^{n} \dfrac{1}{n\sqrt[k]{1+f(n)/n^k}+1}\\
&=\sum_{k=2}^{n} \dfrac1{n}\dfrac{1}{\sqrt[k]{1+f(n)/n^k}+1/n}\\
&\ge\sum_{k=2}^{n} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}\\
\end{array}
$
Since $f(n)/n^c \to 0$
for some $c > 0$,
$f(n) < an^c$
for some $a > 0$
so
$\begin{array}\\
f(n)/(kn^k) 
&\lt an^c/(kn^k)\\
&=a/(kn^{k-c})\\
&\lt a/(kn)
\qquad\text{for } k \ge c+1\\
&\lt 1/n
\qquad\text{for } k > a\\
\end{array}
$
Therefore,
letting
$p(n) = \max(c+1, a)$,
$f(n)/(kn^k) < 1/n$
for $k > p(n)$.
Therefore
$\begin{array}\\
s(n)
&\ge\sum_{k=2}^{n} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}\\
&\ge\sum_{k=2}^{p(n)} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}+\sum_{k=p(n)}^{n} \dfrac1{n}\dfrac{1}{1+f(n)/(kn^k)+1/n}\\
&\gt\sum_{k=p(n)}^{n} \dfrac1{n}\dfrac{1}{1+2/n}\\
&\gt\sum_{k=p(n)}^{n} \dfrac1{n}\dfrac{1}{2}\\
&\gt \frac12\ln(n/p(n))\\
&\to \infty
\qquad\text{since }n/p(n) \to \infty\\
\end{array}
$
