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 $X_n<Y_n$.

Mathematics Asked on January 5, 2022

Question: Let $$x_k, y_kgeq 0$$. Suppose that $$sum_{k=1}^infty x_k and $$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.

My thoughts: So, $$x_k$$ is a convergent sum and $$y_k$$ is a divergent sum both of non-negative terms. Since $$sum_{k=1}^infty x_k, we know that $$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 , $$sum_{k=1}^infty x_k and $$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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