I try to find an equivalent of $S_n=\sum \limits_{k=3}^{n}\frac{1}{k.\log(k).\log(\log(k))}$.
I know that, according to the condensation test with $p=2$, this positive series has the same behavior as $\sum \limits_{n\ge 2}\frac{1}{n.\log(2).\log(n.\log(2))}$ which is divergent by Bertrand's criterion.
Then I tried to find an equivalent of $S_n '=\sum \limits_{k = 2}^n\frac{1}{k.\log(2).\log(k.\log(2))}$ using the integral's test and found that $S_n ' \sim \log(\log(\log(2^n)))$. But apparently $S_n\sim\log(\log(\log(n))).$
I know that $\frac{1}{n.\log(2).\log(n.\log(2))}\not \sim \frac{1}{n.\log(n).\log(\log(n))}$ but it does not imply that $S_n ' \not \sim S_n$. So why can't we find the same result ?
Thanks in advance !
$$\int_3^{n+1}\frac1{x\log x\log\log x}\,dx\le S_n\le\frac1{3\log 3\log \log 3}+ \int_3^n\frac1{x\log x\log\log x}\,dx$$ Therefore, $$ S_n\sim \int_3^n \frac1{x\log x\log\log x}\,dx $$
If you want to use the Cauchy condensation trick to find an equivalent for the partial sums, you have to be careful. The logic behind the condensation test is $$ a_1+(a_2+a_3)+(a_4+a_5+a_6+a_7)+\dots\le a_1+2a_2+4a_4+\dots,\\ \implies \sum_{k=1}^\infty a_k\le \sum_{k=1}^\infty 2^ka_{2^k} $$ So the first three terms of the condensed series account for the first seven terms of the original. In general, the first $k$ terms of the condensed series account for the first $2^k-1$ terms of the original. You can also show on the other end that $$ \sum_{k=1}^\infty 2^ka_{2^k}\le 2\sum_{k=1}^\infty a_k $$
Now, letting $a_k=1/(k\log k\log \log k)$, so that $$ S_n=\sum_{k=3}^n a_{k} $$ $$ S_m'=\sum_{k=3}^m a_{k} $$ then the same trick use in proving the Cauchy condensation test for the whole series implies the following inequalities for the partial sums: $$ S_{2^n+1}\le S'_{n+2}\le 2S_{2^n+1} $$ Note the indice mismatch; the condensed partial sums coerrespond to many terms in the original. This is now consistent with the results $$ S_n\sim \log \log \log n\\ S_n'\sim \log \log \log 2^n $$