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I need to know how to exactly evaluate the following sum (in case the limit exists, which is does I think) :

\begin{equation} \lim_{N \to \infty} \quad \sum_{n = 2}^N \frac{1}{n(n-1)} \end{equation}

The reason I am asking is because I would like know wheter I can manipulate expressions involving series by looking at the sequence of partial sums, so is it true in general that

\begin{equation} \lim_{N \to \infty} \quad \sum_{n = 2}^N (a_n - a_{n-1}) = \lim_{N \to \infty} a_N - a_0 \end{equation}

regardless of the limit behaviour of the whether the sequence $(a_n)$ or the series $\lim_{N \to \infty} \sum^N_{n = 2} a_n$ ?

For example, if I let $S_N = \sum_{n = 1}^N = (-1)^{n+1} \frac{1}{n}$ then the resulting series exists. Writing the same partial sum as \begin{equation} T_N = 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \dots \end{equation} (i.e I rearrange the terms) whe have $t_{3N} = \frac{1}{2} s_{2N}$, hence the second series has half the value of the first, although all we did was rearranging the summands before taking the limit.

So this is where my original quest started - I kind of suspect that in order to do arithmetic manipulations before taking the limit I need make sure the sequence $(a_n)$ has certain properties, right ? If yes, what are these properties ?

Many thks for your help!

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    Those dots in your $T_N$ mean you're really not talking about a finite sum. If you actually picked some $N$ and wrote out $S_N$, you'd see you can't get $T_N$ as there will be terms missing.2012-01-22

2 Answers 2

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Hint: write the summand as a sum of two simpler fractions.

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    thks fo$r$ you answer! So is it true in general that I can manipulate series using the usual rules for sums before taking the limit ? I have edited my question to make it a little bit more clear (hopefully) what I mean by that.2012-01-22
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$\lim_{N \to \infty} \sum_{n=2}^N \dfrac{1}{n(n-1)}=\lim_{N\to \infty}\sum_{n=2}^N\dfrac{n-(n-1)}{n(n-1)}$ So, this boils down to $\lim_{N\to \infty}\sum_{n=2}^N\left(\dfrac{n}{n(n-1)}-\dfrac{n-1}{n(n-1)}\right)\overset{(1)}{=}\lim_{N\to\infty}\left(1-\dfrac{1}{N}\right)=1$

I assume that you know $\left\{\dfrac{1}{n}\right\}_{n=1}^\infty\to0$

The equality (1) comes from the telescoping sum:

$\displaystyle\sum_{n=2}^N(\dfrac{1}{n-1}-\frac{1}{n})=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots-\frac{1}{N-1}+\frac{1}{N-1}-\frac{1}{N}$

Hope this is helpful!

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    many thks for your answer! I edited my question slightly because I am interested in a more general issue concerning manipulations of series.2012-01-22