Crazy Convergence

Let $f_{n}(x)=(\sum_{k=1}^{n}\frac{1}{k^{2}x-1})-\frac{1}{2}$ .

Part 1: $f_{n}(x)$ has a unique solution $x_{n}>1$

$f_{n}(x)$ is continuous in $(1,∞)$ .

See that $lim_{x→1}f_{n}(x)=∞$ and $lim_{x→∞}f_{n}(x)=-\frac{1}{2}$ .

Also $f_{n}'(x)=-\sum_{k=1}^{n}\frac{k^{2}}{(k^{2}x-1)^{2}}<0$ in $(1,∞)$ .

Hence, $f_{n}(x)$ is decreasing in $(1,∞)$ .

So,it is obvious that $f_{n}(x)$ has a unique solution $x_{n}>1$ .

Part 2:The sequence $x_{n}$ converges to $4$ .

$f_{n}(4)=\sum_{k=1}^{n}\frac{1}{4k^{2}-1}-\frac{1}{2}=\frac{1}{2}\sum_{k=1}^{n}(\frac{1}{2k-1}-\frac{1}{2k+1})-\frac{1}{2}=\frac{1}{2}(1-\frac{1}{2n+1})-\frac{1}{2}<0=f(x_{n})$ .

Hence $x_{n}<4$ for all natural numbers $n$ .

$f_{n}(x)$ is differentiable and continuous in $[x_{n},4]$ .

So,by MVT , there exist $t$ belonging to $(x_{n},4)$ such that

$\frac{f(4)-f(x_{n})}{4-x_{n}}=f_{n}'(t)=-\sum\frac{k^{2}}{(k^{2}t-1)^{2}}<-\frac{1}{(t-1)^{2}}<-\frac{1}{(4-1)^2}=-\frac{1}{9}$

Simplify to see that $x_{n}>4-\frac{9}{2n+1}$

So, $4-\frac{9}{2n+1}<x_{n}<4$

Hence, $(x_{n})$ converges to $4$ . :)