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We are basically given that

Our infinite series $\sum a_n$ diverges and $a_n$ is non-negative for each $n$.Does $\sum a_n=+\infty$?

My guess is that since the sequence of partial sums is increasing and divergent, we can conclude by contra-positive statement of Monotone convergence theorem that the sequence of partial sums of $\sum a_n$ is unbounded.

Since every unbounded increasing sequence is properly divergent, our sequence of partial sums shall diverge to $\infty$ and our series shall diverge to $\infty$.

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  • $\begingroup$ Google "Math StackExchange LaTeX guide" for how to write equations. $\endgroup$ – user21820 Mar 8 '16 at 12:30
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More or less correct. But that is called an informal proof, not formal. People who already know how to prove the theorem will understand your explanation, but people who do not will not.

A formal proof would go something like: $\def\nn{\mathbb{N}}$ $\def\rr{\mathbb{R}}$

[Assume that $(a_n)_{n\in\nn}$ is a sequence of real numbers, since it was not stated in the question.]

Let $b_n = \sum_{i=0}^n$, for any $n \in \nn$.

Given any $n\in\nn$:

  $b_{n+1} \ge b_n$ because ...

Therefore $(b_n)_{n\in\nn}$ is increasing.

If $(b_n)_{n\in\nn}$ is bounded above:

  $(b_n)_{n\in\nn}$ has a limit by ... theorem.

  Thus $\sum_{i=0}^\infty$ converges by definition of infinite series.

  Contradiction.

Therefore $(b_n)_{n\in\nn}$ is not bounded above.

Given any $m \in \rr$:

  Let $k \in \nn$ such that $b_k > m$ by definition of boundedness.

  Then $b_n \ge b_k$ for any $n \in \nn_{\ge k}$ because $(b_n)_{n\in\nn}$ is increasing.

  Thus $b_n > m$ for any $n \in \nn_{\ge k}$.

Therefore $(b_n)_{n\in\nn}$ converges to $\infty$ by definition of infinite limit.

Therefore ...

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  • $\begingroup$ Thank you, I'd write the formal version of it in the question. I wasn't really too sure about my answer actually $\endgroup$ – Lelouch Mar 8 '16 at 12:28
  • $\begingroup$ @Lelouch: Oh if you do know how to write a formal version, then you should be able to check with certainty that it is correct, no? $\endgroup$ – user21820 Mar 8 '16 at 12:29
  • $\begingroup$ @Lelouch: See if the formal proof outline I provided helps you to see the logical structure clearly and convince yourself completely of the proof. $\endgroup$ – user21820 Mar 8 '16 at 12:40
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I don't think you really need a contradiction to prove this: having observed the sequence of partial sums $ b_n=\sum_{i=0}^n a_n$ is non-decreasing,

  • Either $ (b_n) $ is bounded, and it converges by the monotone convergence theorem.
  • Or it is unbounded, i.e. for any $A$ there exists a rank $N$ such that $b_N\ge A$. As the sequence is non-decreasing, for all $n\ge N$, $b_n\ge b_N\ge A$, which is the very definition of divergence to $\infty$.
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