# Cannot understand harmonic series logarithmic growth

#### idontknow

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 $$\displaystyle n^2 +2n +1 -n=n^2 +n +1$$?

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#### tahirimanov19

$$\sum_{k=1}^{n^2} 1/k = 2 \ln n + \gamma + \dfrac{1}{2 n^2} + O(( \frac{1}{n} )^4)$$

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