Suppose that $\sum_{j=1}^\infty a_j$ is conditionally convergent. Show that $\sum_{j=1}^\infty j^{\frac{1}{j}}a_j$ is also convergent.

Suppose that $\sum_{j=1}^\infty a_j$ is conditionally convergent. Use the partial sum formula to show that $$\sum_{j=1}^\infty j^{\frac{1}{j}}a_j$$ is also convergent.

Any hints on how to get started?

Let $b_j = j^{1/j}$. The sequence $b_j$ is bounded and decreasing (when $j > 2$) with $b_j \to 1$. The sequence of partial sums $S_n = \sum_{j=1}^n a_j$ is convergent and, hence, bounded.

We have convergence of the telescoping series

$$\sum_{j=1}^n (b_j - b_{j+1}) = b_1 - b_{n+1},$$

I'll leave it to you to show that this implies absolute convergence of

$$\sum_{j=1}^n S_j(b_j - b_{j+1}).$$

Summation by parts (another detail for you) gives

$$\sum_{j=1}^n a_j b_j = S_nb_{n+1} + \sum_{j=1}^nS_j(b_j - b_{j+1}) .$$

Since both terms on the RHS converge so too does the sum on the LHS.

A question for you to consider is where did we use the fact that $b_j$ is eventually decreasing.