User talk:Stefano85

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Comments about User_talk:X42bn6/Working_On/Telescoping_series

My version of telescoping series

I am confused by your comments. What do you mean by not rigorous enough?

However, I am still planning to extend this section. x42bn6 Talk 03:29, 25 January 2006 (UTC)[reply]

A more "rigorous" approach

Resuming what I was talking about:

The series $\sum _{n=1}^{+\infty }{\frac {1}{n^{2}+n}}$ can be represented as:

${\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{12}}+\cdots$

Since the general term of the seqence can be represented (according to the method of Telescoping series) as:

$a_{n}={\frac {1}{n^{2}+n}}={\frac {1}{n}}\cdot {\frac {1}{n+1}}={\frac {1}{n}}-{\frac {1}{n+1}}$

the same series can be represented as:

$\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\left({\frac {1}{3}}-{\frac {1}{4}}\right)+\cdots$

From this representation, one can more easily represent and compute the sequence of partial sums $\{s_{n}\}_{n}$ as: $\{s_{n}\}_{n}\quad =\quad {\frac {1}{2}},\quad {\frac {2}{3}},\quad {\frac {3}{4}},\quad \cdots ,\quad 1-{\frac {1}{n+1}}$ ,

i.e. intuitively it can be said that the general partial sum is $s_{n}=1-{\frac {1}{n+1}}$ .

What I said was not very rigorous is the fact that you infer that the $n$ -th partial sum is $s_{n}=1-{\frac {1}{n+1}}$ , based on the fact that "writing out the first few terms and the last few terms" of the sequence yields that result. The section "A pitfall" gives an example of how applying this method (i.e. rearranging the terms of the series) without the necessary hypotheses of convergence can lead to incorrect results. Before rearranging the terms you'd need to state which hypotheses are verified.

What I was thinking about as an alternative is the following demonstration by mathematical induction:

A demonstration

Basis:

for $n=1$ :

$s_{1}=1-{\frac {1}{n+1}}$

$\left.\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\right|_{{\mbox{for}}~n=1}=1-{\frac {1}{n+1}}$

$1-{\frac {1}{n+1}}=1-{\frac {1}{n+1}}$

Inductive step:

for $s_{n-1}=1-{\frac {1}{n-1+1}}=1-{\frac {1}{n}}$ :

$s_{n}=s_{n-1}+a_{n}=1-{\frac {1}{n}}+{\frac {1}{n}}-{\frac {1}{n+1}}=1-{\frac {1}{n+1}}$

This shows that $\forall n\in N_{0}:\quad s_{n}=1-{\frac {1}{n+1}}$

Only now can we justify that (since given a formula for the general partial sum it is possible to compute the value of the series as its limit for $n\to \infty$ ):

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}+n}}=\lim _{n\to \infty }s_{n}=\lim _{n\to \infty }1-{\frac {1}{n+1}}=1$

Stefano85 23:11, 25 January 2006 (UTC)[reply]

The only real issue that I have with putting this into the article is that fully proving the sums are by induction. Something I don't want to put in the article. As far as I am concerned, the method of differences is still quite a rigorous proof, and in order to incorporate this idea, I would simply put:

This result can also be proved by mathematical induction.

into the article.

Why don't you try editing the article at my userspace itself? Perhaps it would make it slightly easier for me to understand what you are saying (I am only an A-Level student). Alternatively, you could contact User:Michael Hardy or User:Oleg Alexandrov, two of the people who have helped me on this article, and get their views.

And, welcome to Wikipedia. x42bn6 Talk 08:30, 26 January 2006 (UTC)[reply]

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I have saved this article and greatly expanded it. fish&karate 11:28, 16 September 2008 (UTC)[reply]

Ok! Thanks Fish and Karate. I was too busy lately and did't have time to keep an eye on wikipedia.

It's no problem - thank you for the kind words! :) fish&karate 21:30, 19 September 2008 (UTC)[reply]

Associazione Volontari Italiani Sangue

At the time I saw the article on Special:Newpages and tagged it, it was extremely short with no references or citations to secondary sources establishing notability. It's fine now now that it's been improved, and I probably pulled the trigger a little quick. Sorry. TheMolecularMan (talk) 13:41, 19 September 2008 (UTC)[reply]

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