FoxTrot Bill Amend Problems So I found this on the Wolfram website today: 

So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed: 


*

*$ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $

*$ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $ 

*$ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] $ 

*$ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$



Ideas


*

*The degree of the numerator is $\frac{3}{2}$. The degree of the
denominator is $\frac{3}{2}$. The expression is of an indeterminate
form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule. 

*No ideas, really. 

*Derivatives are always pretty easy, although this one is a bit
bashy. Basically bash Product and Chain Rules, etc. 

*Integration by Parts bash? The $\mathrm{d}\theta$ part is very
trivial. I mean for the $\rho$ and $\phi$.
 A: For 1, divide the numerator and denominator by $x^{\frac 32}$  This gets you $\frac {\sqrt{1+\text{small stuff}}}{1+\text{small stuff}}$ which goes to $1$
For 2, Alpha gets a mess:  $\frac 13\left(1-\log 2 + \frac{\pi}{ \cosh \left(\frac {\sqrt 3 \pi}2\right)}\right)$  
For 3, I agree
For 4, the integrals are independent as the limits don't depend on the variables, so it is $\left(\int_0^{2\pi} d\theta \right)\left( \int_0^{\frac \pi 4} \sin \phi \cos \phi d\phi  \right)\left( \int_0^4 \rho^3 d\rho \right)$, all of which are not too tough.
A: My take on $\#3$: $$ \begin {align*} \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] &= \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot (n+1) \ln u - \dfrac {u^{n+1}}{(n+1)^2} \right] \\&= \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1} \ln u}{n+1} - \dfrac {1}{(n+1)^2} \cdot u^{n+1} \right] \\&= \dfrac {1}{n+1} \cdot \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ u^{n+1} \ln u - \dfrac {u^{n+1}}{n+1} \right] \\&= \dfrac {1}{n+1} \cdot \left[ \dfrac {\mathrm{d}}{\mathrm{d}u} \left( u^{n+1} \ln u \right) - \dfrac {\mathrm{d}}{\mathrm{d}u} \left( \dfrac {u^{n+1}}{n+1} \right) \right] \\&= \dfrac {1}{n+1} \cdot \left[ \dfrac {u^{n+1}}{u} + (n+1) \cdot u^n \ln u - u^n \right] \\&= \dfrac {1}{n+1} \cdot \left[ {u^n} + (n+1) \cdot u^n \ln u {-u^n} \right] \\&= \boxed {u^n \ln u}. \end {align*} $$Beautifully done intentionally?
A: I can derive the sum quite easily, using partial fractions and the residue theorem.
Note that $$\frac{k^2}{k^3+1} = \frac13 \left [\frac{1}{k+1} + \frac{2 k-1}{k^2-k+1}\right ]$$
Now, note that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} = 1-\log{2}$$
For the other piece, note the fortuitous coincidence that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = -\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1}$$
(One may see this by showing that the map $k \mapsto -k$ produces the $k+1$th term of the summand.)
Now, the sum on the RHS may be evaluated using residue theory.  In general, one may show that
$$\sum_{k=-\infty}^{\infty} (-1)^k f(k) = -\pi \sum_n \operatorname*{Res}_{z=z_n} [\csc{\pi z} \, f(z)]$$
where the $z_n$ are the non-integral poles of $f$.  In this case, the poles are at $z=e^{\pm i \pi/3}$, and we are also lucky to have an $f$ of the form $g'/g$, so that the residue at each pole is simply $1$.  Therefore, we have
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = \frac{\pi}{2} \left ( \csc{\left (\pi e^{i \pi/3}\right )}+ \csc{\left (\pi e^{-i \pi/3}\right )} \right ) = \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}$$
Therefore,
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} k^2}{k^3+1} = \frac13 \cdot \left [ 1-\log{2} + \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}\right ]$$
