Is My Mathematical Proof Correct?

[Artusartos]

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

Thanks in advance

 

[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.

 

[Artusartos]

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.

Ok, thanks a lot

 

[Artusartos]

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.

Question 10.6 in this link:
http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw3sum06.pdf

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| &lt; 1/n. And I was a bit confused...because can't I use the same proof for this one too?

Thanks in advance

 

[pasmith]

Unfortunately your present proof doesn't work. Your assumption is that |s_{n+k} - s_n| &lt; \epsilon, and you know that |s_{n+ k +1} - s_{n+k}| &lt; 2^{-(n+k)}. The inductive step is then
<br /> |s_{n+k + 1} - s_n| \leq |s_{n+k+1}-s_{n+k}| + |s_{n+k} - s_n| &lt; 2^{-(n+k)} + \epsilon<br />
But, sadly, 2^{-(n+k)} + \epsilon &gt; \epsilon for all n and k. So you can't conclude that |s_{n+k+1}- s_n| &lt; \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.)

The best proof is probably the one given.