$$\sum _{i=1}^{\infty}\frac 1 k e_k$$
where $e_k = \underbrace{(0,\ldots,1}_k,0,\ldots )$,
Prove that series converges to sequence $(1,\frac{1}{2},\frac{1}{3},\ldots)$ in $C_0$.
I know that since $\frac{1}{k}e_k \in C_0 \Rightarrow \lim_k \frac{1}{k}e_k = 0$.
And also $\|\frac{1}{k}e_k\| = 1$.
While looking at series, it is obvious that
$$\sum _{i=1}^\infty \frac 1 k e_k = (1,0,0,\ldots) + \left( 0, \frac 1 2, 0, 0, \ldots \right) + (0,0,\frac{1}{3},0,\ldots) + \cdots$$
But I don't have enough time to write infinite amount of terms and then add them together.
What can be used to reasonably show this addition ? Cauchy sequences perhaps, since it was taught in our course ?
The notation $\displaystyle \sum_{i\,=\,1}^n \frac 1 k e_k$ rather than $\displaystyle \sum_{k\,=\,1}^n \frac 1 k e_k$ or $\displaystyle \sum_{i\,=\,1}^n \frac 1 i e_i$ seems unfortunate.
Since you wrote an expression involving a norm, you have in mind some particular norm, although you did not tell us which one. So you need to show that $$ \left\| \lim_{n\,\to\,\infty} \left( \left( \sum_{k\,=\,1}^n \frac 1 k e_k\right) - \left( 1, \frac 1 2, \frac 1 3, \ldots \right) \right)\right\| =0. $$ Observe that \begin{align} & \left( \sum_{k\,=\,1}^n \frac 1 k e_k \right) - \left( 1, \frac 1 2, \frac 1 3, \ldots \right) \\[10pt] = {} & \Big( \,\underbrace{0,\ldots,0}_n, \frac 1 {n+1}, \frac 1 {n+2}, \frac 1 {n+3}, \ldots \Big) \end{align} So you need to say something about the norm of that last vector. The norm is a positive number. So you have an infinite sequence of positive numbers. The problem is to show that the limit of that sequence of positive numbers is $0.$