How should I decide if the following sequence is convergent or divergent?

$$a_n = \frac{\sum\limits_{k=1}^{n} \frac{1}{k}}{\log n}$$ I would appreciate any approach. Thanks.

I was misunderstood series and sequence so I edited the post accordingly.

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    $\begingroup$ You're asked to decide if a certain sequence converges. The Ratio Test applies to series. $\endgroup$ – Matthew Leingang Sep 22 '17 at 15:01
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    $\begingroup$ By looking at the graph of $yx=1$, you can check that $$\log n \leqslant \sum_{k=1}^n \frac 1k \leqslant 1+\log n $$ for each $n\geqslant 1$, so that the limit of the sequence $a_n$ is $1$. $\endgroup$ – Pedro Tamaroff Sep 22 '17 at 15:03
  • $\begingroup$ The term test is sufficient... $\endgroup$ – Simply Beautiful Art Sep 22 '17 at 15:03
  • $\begingroup$ Are you really being asked if $\sum a_n$ converges? Or if the sequence $a_n$ converges? $\endgroup$ – Thomas Andrews Sep 22 '17 at 15:08

By the integral test proof, you know that $$ \int_1^{n+1}\frac{dx}{x}\leq\sum_{k=1}^n\frac{1}{k}\leq 1+\int_1^{n}\frac{dx}{x} $$ Since $\int\frac{dx}{x}=\ln(x)+C$, you can calculate that the limit converges by the squeeze theorem.


You can use comparison with integrals. Since $f(x)=1/x$ is monotone, $$ \sum_{k=1}^n \frac 1k \ge \int_1^{n+1} \frac 1x dx =\log (n+1) $$ and similarly $$ \sum_{k=1}^n \frac 1k \le 1 + \int_1^n \frac 1xdx =1+\log n $$ Therefore $a_n$ is between $\log(n+1)/\log n$ and $1+1/\log n$ and must converge to 1.

Actually you can show a stronger result: the difference between the $\log n$ and the sum converges to a constant, called Euler–Mascheroni constant.


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