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_nQuestion: Let $x_k, y_k\geq 0$.  Suppose that $\sum_{k=1}^\infty x_k<\infty$ and $\sum_{k=1}^\infty y_k=\infty$.  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 $X_n<Y_n$.
My thoughts: So, $x_k$ is a convergent sum and $y_k$ is a divergent sum both of non-negative terms.  Since $\sum_{k=1}^\infty x_k<\infty$, we know that $S_{x_k}=\sum_{k=1}^n x_k$, the sequence of partial sums is convergent, and since $\sum_{k=1}^\infty y_k=\infty$, we know that $S_{y_k}=\sum_{k=1}^\infty y_k$ is also divergent (to $\infty$ since all terms are non-negative).  But here is where I get stuck.  I'm a bit stuck on how to deal with the $k$ in $X_n$ and $Y_n$, because I can't just pull it out of each series since it's value depends on the sum.  I was thinking that maybe there was a more measure-theoretic way of dealing with this, but I'm not sure.  Maybe it can just be salvaged by dealing with the series and their partial sums?
 A: 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} $    .
A: 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 $X_k < Y_k$.
