# Proof of Convergence

• Dec 3rd 2006, 04:45 PM
fgn
Proof of Convergence
Prove that $\displaystyle \sum_{n=1}^\infty\!a_n$ converges if and only if $\displaystyle \sum_{n=N}^\infty\!a_n$ converges for any $\displaystyle N \geq 1$

My approach..
Assume that the latter series converges then this inequality holds for some real $\displaystyle M$
$\displaystyle \left| \sum_{n=1}^\infty\!a_n - \sum_{n=N}^\infty\!a_n \right| < M$ for any $\displaystyle N \geq 1$

And thus does also the first series converge. They only differ by a constant depending on $\displaystyle N$.

Now assume that $\displaystyle \sum_{n=N}^\infty\!a_n$ diverges to $\displaystyle \pm\infty$ for some $\displaystyle N > 1$ and that $\displaystyle \sum_{n=1}^\infty\!a_n$ converges.

$\displaystyle \sum_{n=1}^\infty\!a_n = \sum_{n=1}^{N-1}a_n + \sum_{n=N}^\infty\!a_n$

This inequality holds for some $\displaystyle M$
$\displaystyle \left| \sum_{n=1}^\infty\!a_n - \sum_{n=1}^{N-1}\!a_n \right| < M$ for any $\displaystyle N \geq 1$
But this inequality doesn't hold for any $\displaystyle M$
$\displaystyle \left| \sum_{n=N}^\infty\!a_n \right| < M$ for any $\displaystyle N \geq 1$
My assumtition must be wrong.
Either does both diverge to $\displaystyle \pm\infty$ or converge.

Is there something wrong? Something missing to complete the proof?
• Dec 3rd 2006, 07:03 PM
ThePerfectHacker
Quote:

Originally Posted by fgn

Is there something wrong? Something missing to complete the proof?

I did not work on the proof but my trianed eagle eye konstantly catches mistakes. You approached this by contradiction. You said "Assume this does not converge...." Then it diverges to $\displaystyle \pm \infty$. That is not true. The sequence $\displaystyle \{(-1)^n\}$ does not converge neither does it diverge to $\displaystyle \pm \infty$
• Dec 4th 2006, 03:51 AM
fgn
Quote:

Originally Posted by ThePerfectHacker
The sequence $\displaystyle \{(-1)^n\}$ does not converge neither does it diverge to $\displaystyle \pm \infty$

You're right. That case botherd me a bit..
But if I divided it into two cases, one case that diverge to $\displaystyle \pm\infty$ and one diverge like your example. I.e the partial sum is oscillating between some value. Then it should be trival to show that your example diverge for any starting index.
• Dec 4th 2006, 10:29 AM
Plato
There is really no need for cases.
Remember that series convergence is about convergence of sequences of partial sums. There is a nice results about sequences: If $\displaystyle \left( {c_n } \right)$ is a sequence and r is number then $\displaystyle \left( {c_n } \right)$ converges if and only if the sequence $\displaystyle \left( {r + c_n } \right)$ converges.

Realizing that $\displaystyle \pm \sum\limits_{n = 1}^N {a_n }$ is a number the result follows quickly.

This result is usually summarized by the old saying: Series convergence is completely determined by what happens to its tail.