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I was fairly easily able to obtain and prove this formula for the sum: $$S(n)=\frac{n}{n+1}$$ by typical means of computing the partial sums, observing the pattern, and proving by induction.

My question is: Is there a more "analytical" means by which to determine this formula?


I ask the same of this question, though let me know if it is too unrelated and should be a separate post:
For $n \in \mathbb{N}$ and $n \geq 2$, find and prove a formula for $\prod_{i=2}^n (1-\frac{1}{i^2})$.

bwllmsn
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1 Answers1

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You can observe that

$$\frac1{i(i+1)}=\frac1i-\frac1{i+1}\;,$$

so the series telescopes:

$$\sum_{i=1}^n\frac1{i(i+1)}=\sum_{i=1}^n\left(\frac1i-\frac1{i+1}\right)=\frac11-\frac1{n+1}=\frac{n}{n+1}\;.$$

For the second problem,

$$\prod_{i=2}^n\left(1-\frac1{i^2}\right)=\prod_{i=2}^n\frac{i^2-1}{i^2}=\prod_{i=2}^n\frac{(i-1)(i+1)}{i^2}=\prod_{i=2}^n\left(\frac{i-1}i\cdot\frac{i+1}i\right)\;,$$

and if you look closely at that last product, you’ll see that the numerator is

$$1\cdot 2\cdot\left(\prod_{k=3}^{n-1}k^2\right)n(n+1)=\frac12(n-1)!(n+1)!\;,$$

while the denominator is $n!^2$. Thus, the whole thing simplifies to $\dfrac{n+1}{2n}$.

Brian M. Scott
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  • Alright, that definitely makes sense. After seeing a form of determining the sum: $\sum_{i=1}^n i$ that was similar to Gauss's version, though using manipulation of indices, allowed the sums to be kept in summation notation, I wondered if there was a similar means for obtaining this. Any attempts to do so ended up getting quickly convoluted. – bwllmsn Sep 20 '15 at 23:40
  • That's exactly what I was looking for in regards to the product question. The "partial products pattern recognition" method was far less obvious than in the summation problem. Thank you again! – bwllmsn Sep 20 '15 at 23:46
  • @Brian: You’re welcome! – Brian M. Scott Sep 20 '15 at 23:49