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This seems to me a trivial question. How many terms does the following sum contain? (This is a quantum sl($N$) invariant) $$ q^{-N+1}+ \ldots + q^{N+1} $$ I am almost sure the answer is $2N+1$ and not $N$. The reason is the following. I can write the previous sum as $$ q^{-N+1}+ \ldots + q^{N+1}= q^{-N+1}+ q^{-N+2}+ \ldots + q^{-N+2N+1} $$ and divide (or not, it does not matter) by $q^{-N}$ to be left with $$ q^{1}+ \ldots + q^{2N+1} $$ and we see that the powers must label the terms. If we add all of them we add $2N+1$ terms.

I want to know if I am crazy or not! In a video about knot invariants Gukov says that this sum has $N$ terms! Later he says we do this in steps of two. But this still does not work since for $N=3$ we have to start with the term $q^1$ and end at the term $q^{7}$ since $7 = 2 \cdot 3 + 1$. I.e. $$ q^1 + q^3 + q^5 + q^7 $$ which gives $N+1$ terms and not $N$ terms. If we do the same for any $N$ we still get $N+1$ terms. E.g. $N=5$ gives 6 terms.

What is the real deal about this quantum sl$(N)$ invariant?

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    In the title, the indexing starts from $-N-1$, not from $-N+1$ according to the analysis you have made. – 2017-02-25
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    I apologise for the typo. The analysis is correct. Do you have any comment on this? – 2017-02-25

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The quantum integer $[N]_q$ is $$[N]_q=\frac{q^N-q^{-N}}{q-q^{-1}}=q^{-N+1}+q^{-N+3}+\dots+q^{N-3}+q^{N-1}$$ It has $N$ terms: using your counting technique, multiplying by $q^{N-1}$ gives $q^0+q^2+\dots+q^{2n-2}$, and $1+\frac{2N-2}{2}=N$.