Given that , $\sum_{k=1}^\infty x_k<\infty$ and $\sum_{k=1}^\infty y_k=\infty$,
Now , let, $ S_{n} = \sum_{k=1}^{n} x_k $ and $ T_{n} = \sum_{k=1}^{n} y_k$,. If possible let, $ X_{n} > Y_{n} $ for all $n $ ,then it implies , $\sum_{k=1}^{n} k x_k > \sum_{k=1}^{n} k y_k $,
this implies $ S_{n} + (S_{n} - S_{n-1}) + (S_{n} - S_{n-2}) + ....... + (S_{n} - S_{2}) + (S_{n} - S_{1}) > T_{n} + (T_{n} - T_{n-1}) + (T_{n} - T_{n-2}) +........ + (T_{n} - T_{2}) + (T_{n} - T_{1}) $.
Now since, $(S_{n}) $ is convergent sequence , so, $(S_{n}) $ is a Cauchy sequence.
Then by definition of Cauchy sequence, there exists natural number $ l $ such that $ |S_{m} - S_{n}| < \epsilon $ for all $ m,n \ge l $ with $m>n $. But $(T_{n}) $ is a divergent sequence , diverge to $\infty $.
So,now for sufficiently large $n $, For left hand side expression $ S_{n} + (S_{n} - S_{n-1}) + (S_{n} - S_{n-2}) + ....... + (S_{n} - S_{2}) + (S_{n} - S_{1}) $
except finitely many of terms , all others terms becomes negligible (by the definition of Cauchy sequence $ S_{n} $) .
But for right hand side expression $ T_{n} + (T_{n} - T_{n-1}) + (T_{n} - T_{n-2}) +........ + (T_{n} - T_{2}) + (T_{n} - T_{1}) $
as $T_{n} $ diverge, it becomes sufficiently larger enough that it could be greater than the left hand side expression.
So, our assumption $ X_{n} > Y_{n} $ goes wrong.
Hence , there exists $n $ such that $ X_{n} < Y_{n} $ .
Best Answer
Since $\sum_{k=1}^{\infty} x_{k}$ is not finite, it is possible to find blocks of integers, say $[n_{i},n_{l+1}]$ for which $\sum_{k=n_{l}}^{n_{l+1} -1} x_k >= 1$ and such that $n_{l+1} > n_{l}$.
For $k\in[n_{l},n_{l+1} -1]$ define $y_{k} = \frac{1}{l}*x_{k}$. Then $$\sum_{k=1}^{\infty} y_{k} = \sum_{l}\sum_{k = n_{l}}^{n_{l+1}-1} y_{k} >= \sum_{l}\frac{1}{l}\sum_{k=n_{l}}^{n_{l+1}-1}x_{k} >= \sum_{l}\frac{1}{l}$$