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In this proof Prove that there exists a sequence $(x_n)$ such that $\sum_n a_n x_n$ diverges, they have assumed $a_n \to \infty$. Here it is not given. So how to find a sequence for the series like $\sum \frac{1}{n}$. How to proceed for this question.

Madhan Kumar
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For $\sum a_n$ to diverge, the sum of the positive terms of $a_n$ must diverge or the sum of the negative terms of $a_n$ must diverge (if both series converged, $\sum a_n$ would as well). So, wlog say that the positive terms diverge. By letting $x_n = 0$ whenever $a_n \leq 0,$ we can in fact assume that $a_n$ only consists of positive terms; so wlog $a_n > 0$ and $\sum a_n = \infty.$

Now, the idea is a lot clearer. We can find a sequence $N_1 < N_2 < \cdots$ so that for each positive integer $k,$ $$\sum_{n=N_{k-1}+1}^{N_k} a_n > k.$$ This is because if you remove finitely many terms from a series diverging to positive infinity, the remaining series still diverges to positive infinity.

Pick $x_n = 1$ for $1 \leq n \leq N_1.$ Thus, $\sum_{n=1}^{N_1} a_nx_n > 1.$

Now, pick $x_n = 1/2$ for $N_1 < n \leq N_2.$ Thus, $$\sum_{n=1}^{N_2} a_nx_n = \sum_{n=1}^{N_1} x_n + \frac{1}{2}\sum_{n=N_1+1}^{N_2} x_n > 1 + 1 = 2.$$

Now do you see what to do? Generally, pick $x_n = 1/k$ for $N_{k-1} < n \leq N_k.$ Then $$\sum_{n=1}^{N_k} a_nx_n > 1 + \frac{2}{2} + \cdots + \frac{k}{k} = k.$$ Hence, $\sum a_nx_n$ still diverges.