Cannot understand harmonic series logarithmic growth

Dec 2015
I know that \(\displaystyle 1+1/2+1/3+...+1/n =\ln(n)+\gamma+\epsilon_{k}=f(n)\). where \(\displaystyle \epsilon_{k}\approx 1/2n\).

The problem is this sum : \(\displaystyle 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n^2 }\)=? , can we go like \(\displaystyle H_{n^2 } =f(n^2 )\)?

If yes then (*) \(\displaystyle \: \displaystyle 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n^2 }=\ln(n^2 )+\gamma +\epsilon(n^2 )\).
Im trying to solve the limit \(\displaystyle \displaystyle l=\lim_{n\rightarrow \infty } \frac{\sum_{i=1}^{n^2 }i^{-1}}{\ln(n)}\) using (*) , since Cesaro-stolz theorem fails.
Now the limit must be \(\displaystyle \lim_{n\rightarrow \infty} \frac{f(n^2 )}{\ln(n)}=2\).

cesaro-stolz theorem cannot fail if we know the number of elements of the difference (which is a sum too) \(\displaystyle d_n =H_{n^2 +2n+1} -H_{n}\).

For example , n=3 , number of elements=16-3=13 ; n=2 , number of elements=7...etc . Can we express \(\displaystyle d_n \) knowing that the number of elements is \(\displaystyle n^2 +2n +1 -n=n^2 +n +1\)?
Last edited:
Mar 2015
Universe 2.71828i3.14159
$$\sum_{k=1}^{n^2} 1/k = 2 \ln n + \gamma + \dfrac{1}{2 n^2} + O(( \frac{1}{n} )^4)
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