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Let $\left\{y_n\right\}$ be a non-decreasing, unbounded sequence. If $a_n:=\sum_{k=1}^{n}(y_n-y_k)$, show (or exhibit a counterexample) that $\lim_{n\to \infty}a_n=+\infty$. I don't have many ideas on how to attack this problem. I calculated $a_{n+1}-a_n = n(y_{n+1}-y_n)$, which shows that $a_n$ is increasing so the limit exists in $\mathbb{R}\cup\left\{+\infty\right\}$. I tried to use $$a_{2n} \geq n(y_{2n}-y_n) $$ But still I can't prove that $n(y_{2n}-y_n)\to \infty$. I have no condition on the growth of $y_{2n}-y_n$ - it could even go to $0$. For instance, the sequence $y_{n}=\sum_{k=1}^{{\left\lfloor \log_2 n\right\rfloor}}\frac{1}{k}$ has $$y_{2n}-y_n = \frac{1}{{\left\lfloor \log_2 n\right\rfloor}+1}\to 0$$ Even though $n(y_{2n}-y_n)\to +\infty$, and $y_n\to \infty$.

I know the idea should be that since $y_n$ blows up, the $y_k$ from $k=1,\dots,n-1$ can't 'keep up' with this growth so $a_n$ blows up as well.

I am convinced that the statement is true because it holds even for slowly increasing sequences such as $\log \log n$.

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$$a_n \ge \sum_{k=1}^{n-1} (y_{k+1} - y_k) = y_n - y_1$$

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Take $y_n=n $

$$a_n=\sum_{k=1}^n (n-k) $$ $$=n^2-\frac {n (n+1)}{2} $$ $$=\frac {n (n-1)}{2} $$

which $\to +\infty $.

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  • $\begingroup$ But I wanted to prove this for a general increasing and unbounded sequence $\left\{y_n\right\}$. $\endgroup$ – Lorenzo Quarisa Aug 15 '17 at 9:58

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