Mathematics Asked on January 5, 2022

Question: Let $x_k, y_kgeq 0$. 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?

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 $X_k < Y_k$.

Answered by cha21 on January 5, 2022

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} $ .

Answered by A learner on January 5, 2022

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