Convergence of $\sqrt{\frac{\sum_{i=1}^n (\frac{1}{i} - \frac{1}{i^2})}{log(n)}}$

Question

I am having trouble with a part of a proof that says:

$\sqrt{\frac{\sum_{i=1}^n (\frac{1}{i} - \frac{1}{i^2})}{log(n)}} \rightarrow 1$

As n goes to infinity. I am aware that the log and harmonic series are related by Euler's constant, but i'm not sure if that's relevant here.

Answer 1

Using Stolz theorem once to get rid of the sum we get: $$\lim_{n\to\infty}\frac{\sum_{i=1}^n \frac{1}{i}-\frac{1}{i^2}}{\log(n)} = \lim_{n\to\infty}\frac{\frac{1}{n}-\frac{1}{n^2}}{\log(n)-\log(n-1)} = \lim_{n\to\infty}\frac{\frac{n-1}{n}-\frac{n-1}{n^2}}{\log[(1+\frac{1}{n-1})^{n-1}]} = 1$$

Answer 2

If you enjoy generalized harmonic numbers,$$S_n=\sum_{i=1}^n \left(\frac{1}{i} - \frac{1}{i^2}\right)=H_n-H_n^{(2)}$$ Using the asymptotics $$S_n=\log \left({n}\right)+\gamma-\frac{\pi ^2}{6} +\frac{3}{2 n}+O\left(\frac{1}{n^2}\right)$$ making, using equivalents, $$\sqrt{\frac{\sum_{i=1}^n (\frac{1}{i} - \frac{1}{i^2})}{\log(n)}} \sim\sqrt{1+\frac{\gamma-\frac{\pi ^2}{6}}{\log(n) } }\sim 1+\frac{\gamma-\frac{\pi ^2}{6}}{2\log(n) } $$

Even for rather "small" values of $n$, the approximation is not too bad. For $n=10$, the exact value is $\frac{1}{504} \sqrt{\frac{350339}{\log (10)}}\approx 0.773937$ while the approximation gives $\approx 0.768148$.

Answer 3

It's standard that $\dfrac{\sum_{k=1}^{n}1/k}{\log n} \to 1.$ And we know $\sum_{k=1}^{n}1/k^2$ converges. Thus

$$\frac{\sum_{k=1}^{n}(1/k-1/k^2)}{\log n} = \frac{\sum_{k=1}^{n}1/k}{\log n} - \frac{\sum_{k=1}^{n}1/k^2 }{\log n} \to 1 + 0 =1.$$

Taking the square root leaves the limit at $1.$