Why does $y_{n+1} - y_n \to 0$? 
Let $f(x) = x\ln x$. Let the sequence $\{y_n\}$ such that $f(y_n) = y_n\cdot \ln y_n = n\pi$.
Why does $y_{n+1} - y_n \to 0$?

Obviously $y_n$ diverges, but we need to prove that the terms in the sequence are getting closer and closer.
The derivative of $f(x)$ is $\ln x + 1$ but I don't how it's helping.
 A: $$
f(y_{n+1})-f(y_n)=(n+1)\pi-n\,\pi=\pi.
$$
On the other hand, by the mean value theorem
$$
f(y_{n+1})-f(y_n)=(y_{n+1}-y_n)f'(z_n)\ge(y_{n+1}-y_n)(\ln(y_n)+1).
$$
Thus
$$
y_{n+1}-y_n\le\frac{\pi}{\ln(y_n)+1}.
$$
A: We have $y_n = f^{-1}(n\pi)$.
Since $f(x)$ is increasing towards $+\infty$ (for $x>1$) and its derivative also increases towards $+\infty$, the derivative of its inverse must decrease towards $0$.
Therefore the Mean Value Theorem will allow you to bound the difference between $f^{-1}((n+1)\pi)$ and $f^{-1}(n\pi)$.
A: We look at $$\lim_{n\to\infty}\frac{f(y_{n+1})-f(y_n)}{y_{n+1}-y_n}=\lim_{n\to\infty}\frac{\pi}{y_{n+1}-y_n}$$
Since this is an approximation to the derivative of $f$ (more specific, this is the slope of the line from $(y_n,f(y_n))$ to $(y_{n+1},f(y_{n+1}))$), and we know that the derivative of $f$ is equal to $f'(x)=\log(x)+1$, we know that the above limit goes to infinity. To support this more, note that $f$ is convex (on $\mathbb{R}^+$), so we know that the slope of the line must be more than the derivative of $f$ in the point $y_n$. Now we know
$$\lim_{n\to\infty}\frac{\pi}{y_{n+1}-y_n}=+\infty$$
so we also know
$$\lim_{n\to\infty}{y_{n+1}-y_n}=0$$
I know this is an intuitional proof, but I'm sure you can make this a "real" mathematically correct proof.
Hope this helped!
