# Showing that partial sums diverge to infinity

## Homework Statement

Let $\sum_{n=1}^{\infty}a_n$ be a series with nonnegative terms which diverges, and let $(s_n)$ be the sequence of partial sums. Prove that $\lim_{n\to\infty} s_n = \infty$.

## The Attempt at a Solution

This isn't a difficult problem, but I want to make sure my details are right.

Note that $(s_n)$ is an increasing sequence that is not convergent. By the monotone convergence theorem, we can conclude that $(s_n)$ is unbounded above. Since we have a sequence that is increasing and unbounded above and does not converge, we see that $\lim_{n\to\infty}s_n=\infty$ ☐

Here is my problem: is this enough to show that $\lim_{n\to\infty}s_n=\infty$? In my textbook, the definition of an infinite limit is this: $\forall M\in\mathbb{R}\exists N\in \mathbb{N}\forall n\ge N, s_n>M$. So is what I said, that the sequence is increasing, unbounded above and does not converge, sufficient to show that this definition is true?

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fresh_42
Mentor
And here is my problem: What does that mean?
Let $\sum_{n=1}^{\infty}a_n$ be a series ... which diverges

And here is my problem: What does that mean?
I believe this means that $\forall n\in \mathbb{N}, a_n\ge 0$.

fresh_42
Mentor
I believe this means that $\forall n\in \mathbb{N}, a_n\ge 0$.
No, that's the meaning of "nonnegative". I substituted this part on purpose by ... in my quote. I mean, what is divergence of a series to you? To me, it is exactly the divergence of the sequence of partial sums. Now, if you make this - my definition - into a statement, which has to be proven, then you obviously used something else to define a divergent series. What is it?

No, that's the meaning of "nonnegative". I substituted this part on purpose by ... in my quote. I mean, what is divergence of a series to you? To me, it is exactly the divergence of the sequence of partial sums. Now, if you make this - my definition - into a statement, which has to be proven, then you obviously used something else to define a divergent series. What is it?
An infinite series is convergent iff the partial sums are convergent. An an infinite series is divergent iff the partial sums are divergent.

We know that $(s_n)$ diverges. I guess the idea of the problem is to show that it specifically diverges to infinity.

fresh_42
Mentor
So a divergent series means, that the sequence of partial sums does not converge.
Let me be correct: $S_n$ converges to $L$ means: $\exists \, L\in \mathbb{R} \,\forall\,\varepsilon>0\,\exists \,N_\varepsilon \in \mathbb{N}\,\forall \,n>N_\varepsilon \,:\, |S_n-L|< \varepsilon$
Then divergences is: $\forall \,L \in \mathbb{R}\,\exists\,C_L>0\,\forall\, n\in \mathbb{N}\,\exists \,N_n > n \,:\,|S_{N_n}-L|>C_L\quad (*)$

So for any potential limit $L$ there is a minimum distance $C_L$ where we find a sequence element outside, if we only go far enough with the index.
[I hope this is correct so far.]

Now how can we conclude from $S_n \leq S_{n+1}$ and the given $(*)$ that $\forall M\in\mathbb{R}\exists N_M\in \mathbb{N}\forall n\ge N_M, s_n>M\,?$
[Do you agree so far? It's a bit late here for so many quantors, so I apologize for possible mistakes.]

I would start with: Let $M\in \mathbb{R}$ be arbitrary but fixed. Then take $L:=M$ and see whether the triangle inequality (possibly its left hand side) can be used.

So a divergent series means, that the sequence of partial sums does not converge.
Let me be correct: $S_n$ converges to $L$ means: $\exists \, L\in \mathbb{R} \,\forall\,\varepsilon>0\,\exists \,N_\varepsilon \in \mathbb{N}\,\forall \,n>N_\varepsilon \,:\, |S_n-L|< \varepsilon$
Then divergences is: $\forall \,L \in \mathbb{R}\,\exists\,C_L>0\,\forall\, n\in \mathbb{N}\,\exists \,N_n > n \,:\,|S_{N_n}-L|>C_L\quad (*)$

So for any potential limit $L$ there is a minimum distance $C_L$ where we find a sequence element outside, if we only go far enough with the index.
[I hope this is correct so far.]

Now how can we conclude from $S_n \leq S_{n+1}$ and the given $(*)$ that $\forall M\in\mathbb{R}\exists N_M\in \mathbb{N}\forall n\ge N_M, s_n>M\,?$
[Do you agree so far? It's a bit late here for so many quantors, so I apologize for possible mistakes.]

I would start with: Let $M\in \mathbb{R}$ be arbitrary but fixed. Then take $L:=M$ and see whether the triangle inequality (possibly its left hand side) can be used.
I'm getting lost in the quantifiers. So I assume that after letting $M$ be fixed, we want to produce an $N_M$. Would could this $N_M$ be? Is it at all related to the $N_n$ from the definition of divergence?

fresh_42
Mentor
I'm getting lost in the quantifiers.
Me, too.
So I assume that after letting $M$ be fixed, we want to produce an $N_M$. Would could this $N_M$ be? Is it at all related to the $N_n$ from the definition of divergence?
I only simply reversed all the quantifiers of convergence to obtain divergence, because you defined it as not convergent. So this definition brings them all in, and I'm not sure if I got all "<" correct. The goal is to distinguish between series which diverge due to alternation or other chaotic behavior and those which tend to infinity. This means, given any $M$ we must find an $N_M$ such that $S_n > M$ for all $n > N_M$.

Perhaps we can go without the non-convergence here. Assume $S_{n+1}\geq S_n$ for all $n$, then either it becomes stationary, in which case the series converges, or it grows beyond all limits. So divergence is only used to exclude the first case and we won't have to use any quantifiers.

• Mr Davis 97