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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 $f(x)$?.

math110
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    Using the usual integral trick, your sum is upper bound by $\int_1^\infty \left(\frac{\ln x}{x}\right)^2 ; ; dx = 2$ (according to online Wolfram integrator). – Chip Apr 05 '16 at 08:06
  • Using h'opital's rule, doesn't one get that $\lim_{x \rightarrow \infty}\left(\frac{\ln x}{x}\right)^2=\lim_{x \rightarrow \infty}\frac{\ln x}{x^2}=\lim_{x \rightarrow 0}\frac{x^2}{\ln x}=0$? Was hoping that you can use the concept of sequences to prove it. – Itakura Apr 05 '16 at 08:35
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    @Chip The integrand is not monotone in $[1, \infty]$, so the integral bound would need a little more work. – Macavity Apr 05 '16 at 09:35
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    @Macavity: i stand corrected. In fact, the sum is $\zeta``(2)$, where $\zeta(x)$ is the Riemann zeta function. Wolfram Alpha gives the results $1.9892802342989...$ I doubt this is "middle school" homework...I would like to see an elementary proof . – Chip Apr 06 '16 at 04:55
  • maybe the Euler-Maclaurin formula would be helpful? :https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula – Chip Apr 06 '16 at 09:12
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    Could you explain us a little more the origin of this question. Is it something you have found in a book ? The last equation intrigue me: $\left(\dfrac{\ln{k}}{k}\right)^2=f(k-1)-f(k)$: as you present it, it is an hint that has been given (for obtaining a telescopic sum). Could you confirm it, or does it come from personal attempts ? – Jean Marie Apr 07 '16 at 03:10
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    This is too much complex for middle school in my opinion. – NoChance Apr 13 '16 at 11:35

2 Answers2

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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.)

mjqxxxx
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$f (x)=\left (\frac{\ln (x)}{x} \right )^2 -$-concave and decreases for $x>5$

Therefore:

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

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

Roman83
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  • That's a very fine adjustment to the integral bound, though you need to explain the $-\frac12f(6)$ correction better. Further I doubt this is the middle school approach the OP wanted :( – Macavity Apr 13 '16 at 11:40