Fix some constant $\delta>0$. Consider a harmonic series ${a_l = \dfrac{1}{l}}$.
Define the sequence $\{n_i\}$ as the following: $n_1 = 0$ and $n_i = \inf \left\{k: \sum\limits_{l=n_{i-1}+1}^k a_l \ge \delta \right\}$ for $i\ge 2$.
Clearly $n_i - n_{i-1}$ indicates the amount of terms needed for the sum $\sum\limits_{l=n_{i-1}+1}^{n_i} a_l$ to be greater than $\delta$, and we expect that the number of terms contained in a block sum grows as terms in a harmonic sequence decrease.
My question is: can we give an asymptotic expression for $n_i$ or some approximate expression of $n_i$?
Additional comment: what if $a_l = \dfrac{1}{l^p}$ for some positive integer $p>1$?
It is well known that $$ \log n + \gamma < \sum\limits_{l = 1}^n {\frac{1}{l}} < \log n + \gamma + \frac{1}{{2n}} $$ for any $n\geq 1$ , where $\gamma$ is the Euler$-$Maclaurin constant (see, e.g., R. M. Young, Euler's constant, Math. Gaz. 75, 187$-$190, 1991.). Thus, $$ \sum\limits_{l = n_{i - 1} + 1}^{n_i } {\frac{1}{l}} = \sum\limits_{l = 1}^{n_i } {\frac{1}{l}} - \sum\limits_{l = 1}^{n_{i - 1} } {\frac{1}{l}} > \log n_i - \log n_{i - 1} - \frac{1}{{2n_{i - 1} }}. $$ Hence, $$ \sum\limits_{l = n_{i - 1} + 1}^{n_i } {\frac{1}{l}} >\delta $$ is definitely satisfied if $$ \log n_i - \log n_{i - 1} - \frac{1}{{2n_{i - 1} }} \ge \delta , $$ i.e., if $$ n_i \ge n_{i - 1} e^{\delta + \frac{1}{{2n_{i - 1} }}} . $$ Regarding your additional question, you may proceed similarly by using \begin{align*} \zeta (p) + \frac{1}{{n^p }}&\left( { - \frac{n}{{p - 1}} + \frac{1}{2} - \frac{p}{{12}}\frac{1}{n}} \right) < \sum\limits_{l = 1}^n {\frac{1}{{l^p }}} \\ & < \zeta (p) + \frac{1}{{n^p }}\left( { - \frac{n}{{p - 1}} + \frac{1}{2} - \frac{p}{{12}}\frac{1}{n} + \frac{{p(p + 1)(p + 2)}}{{720}}\frac{1}{{n^3 }}} \right), \end{align*} where $\zeta$ is Riemann's zeta function (see, e.g., my answer here: https://math.stackexchange.com/q/3603164).