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i) $\sum_{n=a}^\infty\frac1{2n(n+1)} \stackrel{a} = \sum_{n=1}^\infty\frac12(\frac1n-\frac1{n+1}) \stackrel{b} =\frac12 $

ii) $ \sum_{n=1}^\infty\frac1{2n^2-\frac12} \stackrel{1}= \sum_{n=1}^\infty\frac12\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right) \stackrel{2}=1 $

I have problems to see why the identities a and 1 holds and how i can evaluate an get b,2. (The examples i) and ii) are independent)

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Note that we have for:

  • Part a) $$\frac{1}{2}[\frac{1}{n(n+1)}] = \frac{1}{2}[\frac{(n+1)-n}{n(n+1)}] = \frac12[\frac{1}{n}-\frac{1}{n+1}]$$ Note that this is a telescoping series.

  • Part b) In a similar manner, $$\frac{1}{2n^2-\frac12} = \frac{2}{4n^2-1} = \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)} = \frac{1}{2n-1}-\frac{1}{2n +1}$$ another telescoping series.

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Because $$\frac{1}{2}\sum_{n=1}^{+\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{n}-\frac{1}{n+1}+...\right)=$$ $$=\frac{1}{2}\lim_{n\rightarrow+\infty}\left(1-\frac{1}{n+1}\right)=\frac{1}{2}.$$ The second is the similar.

It's just a limit of the telescopic sum.

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