Define $X_n=sum_{k=1}^n kx_k$ and $Y_n=sum_{k=1}^n ky_k$. Prove that there exists an $n$ such that $X_n<Y_n$.

Let $S_n = sum_{i = 1}^n x_i$ and $T_n = sum_{i = 1}^n y_i$.

Let $n$ be a potivite integer such that $T_n > S_n$ and let $a_1, a_2, cdots, a_n$ be non-negative real numbers such that for every $k leq n$,

$$ sum_{i = k}^n ka_i = 1$$

Then we have

$$sum_{i = 1}^n a_iX_i = S_n < T_n = sum_{i = 1}^n a_iY_i$$

So there exists a $k leq n$ with $X_k < Y_k$.

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} $ .