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It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(1), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(1). $$

Edit: The term $O(1)$ (which I have from the paper of Carella) is not correct, please see the valuable comment of Peter with the correct term. Nevertheless the value $6/\pi^2$ is reasonably good for the question posed, if $n$ is large.

It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(\log(x)), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(\log(x)). $$

Edit: The term $O(1)$ (which I hade before from the paper of Carella) is not correct, please see the valuable comment of Peter.
The value $6/\pi^2$ is reasonably good for the question posed, if $n$ is large.

It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(1), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(1). $$

Edit: The term $O(1)$ (which I have from the paper of Carella) is not correct, please see the valuable comment of Peter with the correct term.

It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(1), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(1). $$

Edit: The term $O(1)$ (which I have from the paper of Carella) is not correct, please see the valuable comment of Peter with the correct term. Nevertheless the value $6/\pi^2$ is reasonably good for the question posed, if $n$ is large.

It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(1), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(1). $$

Edit: The term $O(1)$ is not correct, see the valuable comment of Jim.

It is well known that $\sum_{k\le x} \phi(k)=\frac{3}{\pi^2}x^2+O(x\log(x))$. In a similar way we obtain $$ \sum_{k\le x} \frac{\phi(k)}{k}=\frac{6}{\pi^2}x+O(1), $$ by using the Moebius function, i.e., $\sum_{k\le x} \frac{\phi(k)}{k}=\sum_{k\le x}\sum_{d\mid k}\frac{\mu(d)}{d}=\sum_{d\le x}\frac{\mu(d)}{d}\lfloor \frac{x}{d}\rfloor$. Then $$ \sum_{k\le x} \frac{\phi(k)}{k}=x\sum_{d\le x}\frac{\mu(d)}{d^2}-\sum_{d\le x}\frac{\mu(d)}{d}((x/d))=\frac{6}{\pi^2}x+O(1). $$

Edit: The term $O(1)$ (which I have from the paper of Carella) is not correct, please see the valuable comment of Peter with the correct term.