To prove this log inequality (middle school) This following Problem is from a book introduce Telescopic Sums,the book introduce  the idea is use identities to write the sum in the form
$$\sum_{k=2}^{n}[F(k)-F(k-1)]$$
then cansel out terms to obtain $F(n)-F(1)$.
Exercise:
How to show that inequality
$$\sum_{k=1}^{n}\left(\dfrac{\ln{k}}{k}\right)^2=\sum_{k=2}^{n}\left(\dfrac{\ln{k}}{k}\right)^2<2$$
idea:
$$\left(\dfrac{\ln{k}}{k}\right)^2=f(k-1)-f(k)$$ How to find $f(x)$?.
 A: First note that
$$
\frac{1}{k^2}\le \frac{1}{k^2-1/4}=\frac{1}{k-1/2}-\frac{1}{k+1/2}.
$$
So over any range of $k$,
$$
\sum_{k=M}^{N}\frac{\log^2 k}{k^2}\le \log^2 N\sum_{k=M}^{N}\left(\frac{1}{k-1/2}-\frac{1}{k+1/2}\right)=\log^2 N \left(\frac{1}{M-1/2}-\frac{1}{N+1/2}\right),
$$
since the sum telescopes.  In particular, for any increasing sequence of indices $1 \le M_0 < M_1 < M_2 < \cdots$, we have
$$
\sum_{k=M_0+1}^{\infty}\frac{\log^2 k}{k^2} = \left(\sum_{k=M_0+1}^{M_1} + \sum_{k=M_1+1}^{M_2}+\cdots\right)\frac{\log^2 k}{k^2}\le \sum_{i=1}^{\infty}\log^2 M_i\left(\frac{1}{M_{i-1}+1/2}-\frac{1}{M_i + 1/2}\right)\\=\frac{\log^2 M_1}{M_0+1/2}+\sum_{i=1}^{\infty}\frac{\log^2 M_{i+1}-\log^2 M_i}{M_i+1/2}.
$$
If we take $M_i=2^i$ for $i\ge 0$, we get the bound
$$
\sum_{k=2}^{\infty}\frac{\log^2 k}{k^2} \le (\log^2 2) \sum_{i=0}^{\infty}\frac{(i+1)^2-i^2}{2^i+1/2}=(\log^2 2)\sum_{i=0}^{\infty}\frac{2i+1}{2^i+1/2}\approx 2.475,
$$
which isn't bad.  Most of the error is at the low end, so treating the first few terms exactly will improve things.  In particular, take $M_i=2^{i+b}$ instead.  Then
$$
\sum_{k=2}^{\infty}\frac{\log^2 k}{k^2} \le \sum_{k=2}^{2^{b}}\frac{\log^2 k}{k^2} +  (\log^2 2) \left(\frac{(b+1)^2}{2^b+1/2} +\sum_{i=b+1}^{\infty}\frac{2i+1}{2^i+1/2}\right).
$$
For $b\ge 10$, this bounds the original sum below $2$.  (As others have pointed out, hardly a middle-school problem... but at least this estimate uses a telescoping sum in some capacity.)
A: $f (x)=\left (\frac{\ln (x)}{x} \right )^2 -$-concave and decreases for $x>5$
Therefore: 

$$\sum\limits_{m=6}^{\infty}f (m)< \int\limits_{5}^{\infty}f (x)dx+\frac{1}{2}\sum\limits_{m=6}^{\infty}f^{'}(m)<\int\limits_{5}^{\infty}f (x)dx-\frac{1}{2}f (6)$$

$$\sum\limits_{m=2}^{\infty}f (m)<f (2)+f (3)+f (4)+f (5)+\int\limits_{5}^{\infty}f (x) dx-\frac{1}{2}f (6)<2$$
