# Is the sequence $\frac{\sum_{k=1}^{n} \frac{1}{k}}{\log n}$ convergent or divergent?

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.

• You're asked to decide if a certain sequence converges. The Ratio Test applies to series. – Matthew Leingang Sep 22 '17 at 15:01
• 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 $1$. – Pedro Tamaroff Sep 22 '17 at 15:03
• The term test is sufficient... – Simply Beautiful Art Sep 22 '17 at 15:03
• Are you really being asked if $\sum a_n$ converges? Or if the sequence $a_n$ converges? – 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.