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?

  • $\begingroup$ $\lim_{n\to\infty}x_n = \infty$ specifically means $+\infty$, not $\pm\infty$. $\endgroup$
    – Arthur
    Dec 9, 2018 at 19:59
  • $\begingroup$ @Arthur I've made an edit, thank you for the notice $\endgroup$
    – roman
    Dec 9, 2018 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, 2018 at 20:03

1 Answer 1


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 $$\sum_{k=1}^{N+m}x_k\geq \sum_{k=1}^{N}x_k+m\epsilon\geq-\left|\sum_{k=1}^{N}x_k\right|+m\epsilon>-\left|\sum_{k=1}^Nx_n\right|+\left(P+\left|\sum_{k=1}^Nx_n\right|\right)=P.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.