# Do you think my answer is correct?

1. Dec 14, 2012

### Artusartos

Do you think my answer is correct? If not, can you tell me why it is wrong?

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2. Dec 14, 2012

### Michael Redei

This looks fine, but you need to slightly re-adjust the inductive step. There you assume that $|s_{n+k}-s_n|<\epsilon$ and then carry on with
$$\begin{eqnarray*} |s_{n+k+1}-s_n| & = & |s_{n+k+1}-s_{n+k}+s_{n+k}-s_n| \\ & \leq & |s_{n+k+1}-s_{n+k}|+|s_{n+k}-s_n| \\ & < & \frac\epsilon2 + \frac\epsilon2. \end{eqnarray*}$$
That last line should be "$\ldots<\epsilon+\epsilon$". Which means you need to go back in your proof and somehow ensure that $|s_{n+k}-s_n|$ and $|s_{n+k+1}-s_{n+k}|$ are together less than $\epsilon$, not each separately.

3. Dec 14, 2012

### Artusartos

Ok, thanks a lot

4. Dec 31, 2012

### Artusartos

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw3sum06.pdf [Broken]

is the same as the one that I solved...

But in part (b), it says that this result is not true for $$|s_{n+1} - s_n| < 1/n$$. And I was a bit confused...because can't I use the same proof for this one too?

Unfortunately your present proof doesn't work. Your assumption is that $|s_{n+k} - s_n| < \epsilon$, and you know that $|s_{n+ k +1} - s_{n+k}| < 2^{-(n+k)}$. The inductive step is then
$$|s_{n+k + 1} - s_n| \leq |s_{n+k+1}-s_{n+k}| + |s_{n+k} - s_n| < 2^{-(n+k)} + \epsilon$$
But, sadly, $2^{-(n+k)} + \epsilon > \epsilon$ for all $n$ and $k$. So you can't conclude that $|s_{n+k+1}- s_n| < \epsilon$, even though this might be the case. (If your inductive step doesn't make use of the given bound on $|s_{n+k +1} - s_{n+k}|$, then your proof is almost certainly flawed.)