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If $b_n = \frac{1}{2n+1}$ does the notation $b_{n+1}$ make $b_n = \frac{1}{2(n+1)+1}$ or $b_n = \frac{1}{2n+2}$?

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What does this have to do with sum notation? – Qiaochu Yuan Jan 20 '11 at 3:34
up vote 4 down vote accepted

The former: $b_{n+1} = \frac{1}{2(n+1)+1}$.

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Ask yourself what you are replacing with $n+1$ to get $b_{n+1}$? You are replacing $n$, right? Therefore you also replace $n$ with $n+1$ in the expression on the right hand side of the equation. This gives $b_{n+1} = \frac{1}{2(n+1)+1}$ as PEV said.

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