# A question concerning the arithmetic manipulation of Series

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

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

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

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

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 $$T_N = 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \dots$$ (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 ?

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About whether you can manipulate like that, the answer is no. The answer supplied by Gerry Myerson expected that you would use the hint to find an explicit formula for the sum of the terms from $2$ to $N$ (for which anything is allowed, it is a finite sum) and then take the limit. –  André Nicolas Jan 22 '12 at 1:47
Ok that makes sense ! I edited the post slightly because I just want to make sure I understand what you say. From what I see you suggest I can always manipulate the partial sum, before taking the limit, without imposing further restrictions on the sequence $(a_n)$. Is that correct ? –  harlekin Jan 22 '12 at 1:57
Before you take the limit, it's a finite sum, and you can manipulate those to your heart's content. –  Gerry Myerson Jan 22 '12 at 1:59
@harlekin: Your modification is just fine. The equality has to be taken in the sense that the limit on the left exists iff the limit on the right exists, and in that case the two sides are equal. –  André Nicolas Jan 22 '12 at 2:05
@Gerry: this makes me a little unsure I understand the example I added to the post above of a series where manipulating the finite sum before taking the limit acutally changes the value of the limit. I took this straight from Victor Bryan's book "Yet another Introduction to Analysis" (page 73). What is the difference between my example and your notion of manitpulating finite sums ? Thks for your help! –  harlekin Jan 22 '12 at 2:13

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

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Partial fractions are not only for integration. –  André Nicolas Jan 22 '12 at 1:28
thks for 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. –  harlekin Jan 22 '12 at 1:33

$$\lim_{N \to \infty} \sum_{n=2}^N \dfrac{1}{n(n-1)}=\lim_{N\to \infty}\sum_{n=2}^N\dfrac{n-(n-1)}{n(n-1)}$$ 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:

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