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October 16th, 2019, 06:15 AM   #1
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Cannot understand harmonic series logarithmic growth

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 by idontknow; October 16th, 2019 at 06:48 AM.
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October 16th, 2019, 12:11 PM   #2
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$$\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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