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Considering something like this:

$ \frac{10}{10} + \frac{9}{11} + \frac{8}{12} + ...$

Where denominator increases each iteration while the numerator decreases. Is there a simple way to find the sum of n iterations of this without doing all the iterations?

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  • $\begingroup$ Do you stop if the numerator hits zero or do you continue? $\endgroup$ – neofoxmulder Feb 18 '14 at 20:38
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$$\sum_{j=0}^{n} {\frac{10-j}{10+j}}$$

$$=-\sum_{j=0}^{n} {\frac{j+10-20}{10+j}}$$

$$=-\sum_{j=0}^{n} {\frac{j+10}{10+j}} + \sum_{j=0}^{n} {\frac{20}{10+j}}$$

$$= -(n+1) + 20 (\sum_{j=1}^{n+10} {\frac{1}{j}}-\sum_{j=1}^{10} {\frac{1}{j}}) $$

$$ = 20H_{n+10} -n -1-20\sum_{j=1}^{10} {\frac{1}{j}}$$

$$ = 20H_{n+10} -n -\frac{7255}{126}$$

Where $H_{k} = \sum_{j=1}^{k} {\frac{1}{j}}$ is the $k$th harmonic number. The harmonic numbers have formulas. Check:

http://mathworld.wolfram.com/HarmonicNumber.html

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  • $\begingroup$ This might not be what the OP meant. I interpreted it as the sum n/n+(n-1)/(n+1)+(n-2)/(n+2)+...+1/(2n-1). $\endgroup$ – Flowers Feb 18 '14 at 21:47
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Using your example: Start by rewriting this as $1+\sum_{n=1}^{\infty}\frac{10-n}{10+n}$ (or you can re-index and start at $0$). I did it this way to show you the lower bound.

Comment: If this series is summable then: $\lim_{\ n\to \infty} \frac{10-n}{10+n} \to 0$. Proof

Does this sequence go tend to $0$?

I say this to say that we need to be specific because as we see arbitrary sequences like this aren't summable.

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$\sum\limits_{k=0}^{n-1} 1-\frac{2k}{n+k}=n-2\sum\limits_{k=1}^{n-1}\frac{k}{n+k}=n-2\sum\limits_{k=1}^{n-1}1-\frac{n}{n+k}= n-2(n-1)+2n\sum\limits_{k=1}^{n-1}\frac{1}{n+k}=\\ 2-n+2n(H_{2n-1}-H_n)$

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