We say that a functional in a normed space attains it's norm if: $$\exists x \in \overline{B}(0,1).|f(X)|=\| f \|$$
For the functionals in $l_1$: $$a_n \mapsto \sum a_n/n$$ and $$a_n \mapsto \sum \Big(1-\frac 1 n\Big)a_n$$ How can I see if they attain their norm?
The first one, as you wrote in the comments, has norm $1$, which is attained at $(1,0,0,0,\ldots)$.
The second one also has norm $1$, because, for each $m\in\mathbb N$, if $(a_1,a_2,\ldots)$ is the sequence such that $a_m=1$ and all others $a_n$'s are equal to $0$, then$$\left|\sum_{n=1}^\infty\left(1-\frac1n\right)a_n\right|=1-\frac1m$$ and therefore the norm is at least $1$. On the other hand, if $(a_1,a_2,\ldots)\in\ell^1$ and it has norm $1$, then$$\left|\sum_{n=1}^\infty\left(1-\frac1n\right)a_n\right|\leqslant\sum_{n=1}^\infty\left(1-\frac1n\right)|a_n|\leqslant\sum_{n=1}^\infty|a_n|\leqslant1.$$However, in this case, the norm is never attained. In fact, if there was a sequence $(a_1,a_2,\ldots)\in\ell^1$ with norm $1$ such that$$\left|\sum_{n=1}^\infty\left(1-\frac1n\right)a_n\right|=1,$$then$$1=\left|\sum_{n=1}^\infty\left(1-\frac1n\right)a_n\right|\leqslant\sum_{n=1}^\infty\left(1-\frac1n\right)|a_n|\leqslant\sum_{n=1}^\infty|a_n|\leqslant1.$$Therefore, all these inequalities are, in fact, equalities. But then\begin{align}0&=\left(\sum_{n=1}^\infty|a_n|\right)-\left(\sum_{n=1}^\infty\left(1-\frac1n\right)|a_n|\right)\\&=\sum_{n=1}^\infty\frac{|a_n|}n\end{align}and therefore $(\forall n\in\mathbb{N}):a_n=0$, which is impossible, since we are assuming that $\bigl\|(a_1,a_2,\ldots)\bigr\|_1=1$.