<s>[INST] Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
 [/INST] You want
$\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
$.
More generally,
consider
$f_{a, c}(n)
=\sqrt[a]{n^a+n^{a-c}}
$
where
$a> 0$
and
$a > c > 0$.
I will use the
generalized binomial theorem
and generalized binomial coefficients.
We have
$\begin{array}\\
f_{a, c}(n)
&=\sqrt[a]{n^a+n^{a-c}}\\
&=n\sqrt[a]{1+n^{-c}}\\
&=n(1+n^{-c})^{1/a}\\
&=n\sum_{k=0}^{\infty} \binom{1/a}{k}n^{-ck}
\qquad\text{(generalized binomial theorem)}\\
&=n\sum_{k=0}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!}n^{-ck}
\qquad\text{(see (**) below)}\\
&=n\left(1+\sum_{k=1}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!n^{ck}}\right)\\
&=n+\sum_{k=1}^{\infty} \dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^{k}k!n^{ck-1}}\\
&=n+\dfrac{1}{an^{c-1}}+\dfrac{1-a}{2a^2n^{2c-1}}+\dfrac{(1-a)(1-2a)}{6a^3n^{3c-1}}+...\\
\end{array}
$
If $c=1$,
$f_{a, 1}(n)
=\sqrt[a]{n^a+n^{a-1}}
=n+\dfrac{1}{a}+\dfrac{1-a}{2a^2n}+\dfrac{(1-a)(1-2a)}{6a^3n^{2}}+...
$.
In particular
$f_{2, 1}(n)
=\sqrt{n^2+n}
=n+\dfrac{1}{2}-\dfrac{1}{8n}+\dfrac{3}{48n^{2}}+...
=n+\dfrac{1}{2}-\dfrac{1}{8n}+\dfrac{1}{16n^{2}}+...
$
and
$f_{3, 1}(n)
=\sqrt[3]{n^3+n^{2}}
=n+\dfrac{1}{3}-\dfrac{2}{18n}+\dfrac{10}{162n^{2}}+...
=n+\dfrac{1}{3}-\dfrac{1}{9n}+\dfrac{5}{81n^{2}}+...
$.
Therefore the limit is
$\lim_{n \to \infty} ((n+\dfrac{1}{2}+O(1/n))-(n+\dfrac{1}{3}+O(1/n)))
=\dfrac16
$.
Note:
In this kind of computation,
I compute first and
worry about convergence later
(or never).
**
We have,
since
$\binom{r}{k}
=\dfrac{\prod_{j=0}^{k-1}(r-j)}{k!}
$,
$\begin{array}\\
\binom{1/a}{k}
&=\dfrac{\prod_{j=0}^{k-1}(\frac1{a}-j)}{k!}\\
&=\dfrac{\prod_{j=0}^{k-1}(1-aj)}{a^kk!}\\
\end{array}
$
</s>