query: Is $\sum^n_{i=1}\sum_{s=1}^{i-1}=\sum_{s=1}^n\sum_{i=s+1}^n$ and in that case why?
positive_passages.docid: 145900
positive_passages.text: Proof of a double summation reduction identity
negative_passages.docid: 47313
negative_passages.text: What is $i^i$ and $(i^i)^i$
negative_passages.docid: 121554
negative_passages.text: Finding $\sum\frac{1}{2^n(n+1)}$
negative_passages.docid: 137967
negative_passages.text: Are there $a,b \in \mathbb{N}$ that ${(\sum_{k=1}^n k)}^a = \sum_{k=1}^n k^b $ beside $2,3$
negative_passages.docid: 89771
negative_passages.text: $\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} - \frac{1}{n^{s-1}}\right)\right]$
negative_passages.docid: 68725
negative_passages.text: Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $
negative_passages.docid: 3798
negative_passages.text: Finding formula for sum $S_n = \sum_{i=1}^n i(i-1)$
negative_passages.docid: 25882
negative_passages.text: Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$
negative_passages.docid: 57780
negative_passages.text: $\sum_{i=1}^n i\cdot i! = (n+1)!-1$ By Induction
negative_passages.docid: 28254
negative_passages.text: Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$?
negative_passages.docid: 99420
negative_passages.text: Compute $\sum_{n=1}^\infty \frac{1}{(n+1)\sqrt[]{n}+n\sqrt[]{n+1}}$
negative_passages.docid: 114575
negative_passages.text: If $\sum_{i=1}^k \dim F_i > n(k-1)$ then $\cap_{i=1}^k F_i \neq \{0\}$
negative_passages.docid: 135927
negative_passages.text: Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$
negative_passages.docid: 143236
negative_passages.text: $ S_{n}=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+...........+\frac{x^{2^{n}}}{(x+1)(x^2+1)...(x^{2^{n}}+1)}$
negative_passages.docid: 89758
negative_passages.text: Is there a formula for $(1+i)^n+(1-i)^n$?
negative_passages.docid: 134311
negative_passages.text: How to determine equation for $\sum_{k=1}^n k^3$
negative_passages.docid: 42216
negative_passages.text: Sum of infinite series $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}$