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For the sequence $a_n =\frac{1}{n}\sum_{i =2}^n \frac{1}{\ln i}$ (with $n \ge 2$), I would like to determine if the limit exists, and, if so, find its value.

Some observations I have made so far: integral comparison does not seem to help--we do have $a_n \le \sum_{i = 2}^n \frac{1}{i\ln i}$. But, by integral comparison, this sum diverges as $n \to \infty$. To get some concrete handle on the problem, I have computed some values in the sequence and found $a_{10} \approx .61$, $a_{100} \approx .3$. So the series appears to be approaching $0$. Is there a standard analysis trick I am missing or is there some more advanced technique needed to establish the limit?

Did
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JZS
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3 Answers3

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Let $b_n\to0$ and, for every $n\geqslant2$, $a_n=\frac1n\sum\limits_{i=2}^nb_i$. Then, for every $\varepsilon\gt0$, $|b_n|\leqslant\varepsilon$ for every $n\geqslant N_\varepsilon$ hence $|a_n|\leqslant\frac1n\left|\sum\limits_{i=2}^{N_\varepsilon-1}b_i\right|+\frac1n\sum\limits_{i=N_\varepsilon}^n|b_i|\leqslant\frac1n\,N_\varepsilon|a_{N_\varepsilon}|+\varepsilon$ and $\limsup\limits_{n\to\infty}|a_n|\leqslant\varepsilon$. Thus, $\lim\limits_{n\to\infty}a_n=0$.

Use this for $b_n=\frac1{\log n}$.

Did
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$$\sum_{j=2}^n\frac{1}{\log j}=\sum_{j=2}^k\frac{1}{\log j}+\sum_{j=k+1}^n\frac{1}{\log j}\leq\dfrac{k}{\log 2}+\dfrac{n-k}{\log k}$$ Now choose $k=[\log n]$ and we find the limit to be $0$.

Ishan Banerjee
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You can use the Abel summation technique from here to derive the asymptotic. We have \begin{align} S_n & = \sum_{k=2}^n \dfrac1{\log(k)} = \int_{2^-}^{n^+} \dfrac{d \lfloor t \rfloor}{\log(t)} = \dfrac{n}{\log(n)} - \dfrac2{\log(2)} + \int_2^{n} \dfrac{dt}{\log^2(t)}\\ & \leq \dfrac{n}{\log(n)} - \dfrac2{\log(2)} + \int_2^{n} \dfrac{dt}{\log^2(t)} =\dfrac{n}{\log(n)} - \dfrac2{\log(2)} + \int_2^{3} \dfrac{dt}{\log^2(t)} + \int_3^{n} \dfrac{dt}{\log^2(t)}\\ &\leq \dfrac{n}{\log(n)} \overbrace{- \dfrac2{\log(2)} + \int_2^{3} \dfrac{dt}{\log^2(t)}}^{\text{constant}} + \dfrac1{\log(3)}\int_3^{n} \underbrace{\dfrac{dt}{\log(t)}}_{\leq S_n}\\ \end{align} We have get that $$S_n \leq \dfrac{n}{\log(n)} + \text{constant} + \dfrac{S_n}{\log(3)} \implies S_n \leq \dfrac{\log(3)}{\log(3)-1} \left(\dfrac{n}{\log(n)} + \text{constant}\right) \tag{$\star$}$$ Hence, we get that $$\dfrac{S_n}n \leq \dfrac{\log(3)}{\log(3)-1} \left(\dfrac1{\log(n)} + \dfrac{\text{constant}}n\right)$$ Now take the limit to get the answer as $0$. You could of course derive better bounds for $(\star)$.

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