TITLE: Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$?
QUESTION [30 upvotes]: I've been struggling for a while with the following problem:
Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$?
Needless to say software aid is not allowed. All manual calculations should be possible to be performed in reasonable time (for example, I did read some solutions to similar problems that to my understanding include calculating 113th power of a decimal number, and this does not count as "reasonable" manual calculation).

Probably useless for the solution, but nevertheless interesting graphs of  $(x+1)^{x+1}-x^{x+2}$:




Using computer:
$(\pi+1)^{\pi+1} = 359.796$
$\pi^{\pi+2} = 359.867$

REPLY [4 votes]: Just to be clear on what it would take to decide this based on rational approximation: we want to compare the ratio
$$
f(x)=\frac{x^x}{(x-1)^{x+1}}
$$
to unity at $x=(\pi + 1)$.  Its logarithm is
$$
\begin{eqnarray}
g(x)=x\log x - (x+1)\log(x-1) &=& x\log x - (x+1)\log x - (x+1)\log(1-1/x) \\
&=& -\log x - (x+1)\log(1-1/x) \\
&=& -\log x + (x+1)\sum_{k=1}^{\infty}\frac{1}{k}x^{-k} \\
&=& -\log x + 1 +\sum_{k=1}^{\infty}\left(\frac{1}{k}+\frac{1}{k+1}\right)x^{-k},
\end{eqnarray}
$$
valid for $x>1$, the derivative of which is
$$
g'(x)=-\frac{1}{x}-\sum_{k=2}^{\infty}\left(1+\frac{k-1}{k}\right)x^{-k},
$$
which is clearly negative.  So $f(x)$ is decreasing, and is less than one at $x=(\pi + 1)$ if it less than one for some rational $(p/q) < \pi+1$.  A good rational underestimate for $\pi$ is given by $\frac{333}{106}\approx 3.14151 < \pi$.  Indeed, one finds numerically that $f(333/106 + 1) =f(439/106) < 1$ by a few percent:
$$
f\left(\frac{439}{106}\right)^{106}=\frac{(439/106)^{439}}{(333/106)^{545}}=\frac{439^{439} 106^{106}}{333^{545}} \approx 0.98.
$$
Despite being integers, though, the numbers involved are too large for manual calculation.  We can do better by looking at the second derivative of $f(x)$: it is $f''(x) = (g'(x)^2 + g''(x))f(x)$, which is clearly positive, since $g''(x)$ is.  Therefore $f(x)$ always lies below any line segment that connects two points on it.  So for any $x < y < (\pi+1) < z$, we have
$$
f(\pi + 1) < f(y) < \lambda f(x) + (1-\lambda) f(z),
$$
where $\lambda = (z-y)/(z-x)$.  This estimate is good enough for $(x,y,z)=\left(4,\frac{439}{106},\frac{29}{7}\right)$, for which $\lambda = 1/106$, $f(x)=256/243$, and $$f(z)=7\sqrt[7]{\frac{29^{29}}{22^{36}}}=\frac{7\cdot 29^{4}}{22^{5}}\sqrt[7]{\frac{29}{22}}.$$  Putting it together, we need to show that
$$
\frac{128}{12879} + \frac{105\cdot 7 \cdot 29^{4}}{106\cdot 22^{5}}\sqrt[7]{\frac{29}{22}} < 1,
$$
or
$$
\sqrt[7]{1+\frac{7}{22}} < \frac{12751\cdot 106 \cdot 22^5}{12879 \cdot 105 \cdot 7 \cdot 29^4}.
$$
Now, the root is less than its fifth-order Taylor series approximation:
$$
\sqrt[7]{1+7x} < 1 + x - 3x^2 + 13x^3 - 65x^4 + 351x^5 = \frac{5361157}{5153632},
$$
evaluated at $x=1/22$.  So in the end it will suffice to show that
$$
{5361157 \cdot 12879 \cdot 105 \cdot 7 \cdot 29^4} < {5153632 \cdot 12751\cdot 106 \cdot 22^5}.
$$
While these are certainly big numbers (about $20$ digits), they're much smaller than the thousand-digit monsters involved in the approach relying only on the first derivative, and I think you could argue that even if you didn't want to, you could carry out this calculation by hand to prove that $(\pi+1)^{\pi+1} < \pi^{\pi+2}$.