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Why is it true that $\sum\limits_{k=2}^{n}{k \over \ln k} \le {n^2 \over \ln n}, n \ge 2$

I try to expand the term of the sum in taylor series but it didn't help me.
I try to recognize the sum as a lower Riemman sum of $\int{x \over lnx}$, but it didn't help either.

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    Hint: On $[e,\infty)$, the function $\frac{x}{\ln x}$ is increasing.2011-11-07

1 Answers 1

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First dispose off the cases $n=2$ and $n=3$ by explicit verification. We will henceforth assume $n \geq 4$.

Since the function $ f(x) = \frac{x}{\ln x} $ is increasing in $[e, \infty)$, it follows that $ \frac{k}{\ln k} \leq \frac{n}{\ln n} \tag{1} $ for $3 \leq k \leq n$. Also, assuming $n \geq 4$, we have $ \frac{2}{\ln 2} = \frac{4}{\ln 4} \leq \frac{n}{\ln n}. \tag{2} $ Adding the inequalities $(1)$ (for $3 \leq k \leq n$) and $(2)$, we get $ \sum_{k=2}^{n} \frac{k}{\ln k} \leq \frac{n(n-1)}{\ln n}, $ which implies the result.

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    Very clear argument. I've learn instantly from it. Thank you.2011-11-09