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Given: $$ \lim_{n\to\infty}x_n = \infty $$ show that $$ y_n = \left\{ \sum_{k=1}^n x_k\right\} $$ is an unbounded sequence.

Intuitively this is obvious, however I'm having a hard time proving that formally.

Start with definition of a divergent sequence we have that for any $\varepsilon > 0$ there exists some $N$ such that $x_n > \varepsilon$: $$ \forall \varepsilon >0 \ \exists N\in\Bbb N: \forall n > N \implies x_n > \varepsilon $$

With that in mind there must be some index starting from which the sequence becomes a monotonically increasing sequence. Now if we consider the difference between sums we may obtain: $$ y_{n} - y_{n-1} = x_n \\ y_{n+1} - y_n = x_{n+1} $$

So from this we (hopefully) may conclude that the difference between the terms of $y_n$ is also increasing which means that the whole sum is also monotonically increasing which means it has no upper bound.

The problem with the above is that it doesn't feel like a formal proof and I have strong doubts about the validity of my reasoning.

Eventually the question is how to prove what's in the problem section in valid formal way?

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  • $\begingroup$ $\lim_{n\to\infty}x_n = \infty$ specifically means $+\infty$, not $\pm\infty$. $\endgroup$ – Arthur Dec 9 '18 at 19:59
  • $\begingroup$ @Arthur I've made an edit, thank you for the notice $\endgroup$ – roman Dec 9 '18 at 20:02
  • $\begingroup$ In the first case $\sum_{k=N+1}^n x_k > \epsilon(n -N) \to +\infty$ as $n \to \infty$ and the sum for $1 \leqslant k \leqslant N$ is fixed. $\endgroup$ – RRL Dec 9 '18 at 20:03
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So you have $x_n>\epsilon>0$ for all $n\geq N.$ Now choose $P\in\mathbb{R}$. By Archimedes choose $m\in\mathbb{N}$ such that $m\epsilon>P+|\sum_{k=1}^Nx_n|$. Then we have $$\sum_{k=1}^{N+m}x_k\geq \sum_{k=1}^{N}x_k+m\epsilon\geq-\left|\sum_{k=1}^{N}x_k\right|+m\epsilon>-\left|\sum_{k=1}^Nx_n\right|+\left(P+\left|\sum_{k=1}^Nx_n\right|\right)=P.$$

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