Vtyjkyuk

Find the value of
$$lim_nrightarrow inftyfrac1+frac12+frac13+cdots +frac1n^3ln(n)$$

My Try: Using Stolz-Cesaro,

Let $displaystyle a_n = 1+frac12+frac13+cdots cdots +frac1n$
and $b_n = ln(n)$

So $displaystyle fraca_n+1-a_nb_n+1-b_n = lim_nrightarrow inftyfrac1(n+1)^3cdot frac1lnbigg(1+frac1nbigg) = 0$

Please explain if what I have done above is right.

Let $H_n=sum_k=1^n frac 1k$. Since $H_n=log n +O(1)$, $$H_n^3 = 3log n +O(1)$$
thus$$fracH_n^3log n = 3+Oleft(frac1log n right)=3+o(1)$$

The limit is $3$.

Use of Cesaro-Stolz is tricky here. Let's first evaluate the limit of $a_n$ given by $$a_n=frac1log n sum_k=1^nfrac1k$$ Using Cesaro-Stolz we have $$lim_nto infty a_n=lim_ntoinfty frac1/nlog n-log(n-1)=-lim_nto infty frac1nlog(1-(1/n))=1$$ and then the sequence in question, say $b_n$, is given by $b_n=3a_n^3$ so $b_nto 3$ as $ntoinfty$.

Your $a_n$ should go up to $frac1n^3$. Then $a_n+1-a_n=frac1(n+1)^3+frac1(n+1)^3-1dots+frac1n^3+2+frac1n^3+1$.

I don't know whether you can arrive there by your method, but you can find the limit by using the asymptotics of harmonic numbers / definition of the euler $gamma$ constant. Then the limit should be $3$.

Note that using Stolz-Cesaro we obtain

$$fracsum_k=1^(n+1)^3frac1k-sum_k=1^n^3frac1kln(n+1)-ln nsimfracln(n+1)^3-log n^3ln(1+frac1n)=3fracln(1+frac1n)ln(1+frac1n)=3$$

but, as already noted by Gabriel Romon, this is not necessary and it is sufficient use that by Harmonic series

$$fracsum_k=1^n^3frac1kln nsimfracln n^3ln n=3$$

To answer your question directly. The mistake is made in $a_n+1-a_n$. For example, if $n = 10$, then this is given by
$$a_11-a_10 = 1/1331 + 1/1330 + cdots + 1/1001$$
since $11^3 = 1331$ and $10^3=1000$.

If you insert this, you still are going to need to do something similar as Gabriel suggested by viewing the summation as upper and lower Riemann sums.

Edit2: if you now use Stolz Cesaro, which is possible, it works out as Gimusi did.