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infinite_geometric_series [2016/07/10 22:01] nikolaj old revision restored (2016/07/10 17:44) |
infinite_geometric_series [2016/07/10 22:33] nikolaj |
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| $ \sum_{k=0}^{n-1} \sum_{m=0}^k {k \choose m}(-z)^{-m} = z\left(1-\left(\dfrac{z-1}{z}\right)^n\right)$ | $ \sum_{k=0}^{n-1} \sum_{m=0}^k {k \choose m}(-z)^{-m} = z\left(1-\left(\dfrac{z-1}{z}\right)^n\right)$ | ||
| - | We can get rid of the double sum as follows: | + | See also [[Niemand series]]. |
| - | + | ||
| - | $ = \sum_{k=0}^{n-1}\left(\prod_{j=0}^k\dfrac{n-j}{1+j}\right)(-z)^{-k} $ | + | |
| - | + | ||
| - | But note that here the summands depends on the upper bound of the sum and for this we can't take the limit $n\to \infty$ of the summand term anymore. | + | |
| - | + | ||
| - | There are some related sums, e.g. | + | |
| - | + | ||
| - | <code> | + | |
| - | expFactor[n_] = (n - (k - j))/(j*f[n]); | + | |
| - | invFactor[n_] = -(n - j)/(j + 1); | + | |
| - | Sum[Product[expFactor[n]*z, {j, 1, k}], {k, 0, n}] // Simplify | + | |
| - | Sum[Product[invFactor[n]*z, {j, 1, k}], {k, 0, n}] // Simplify | + | |
| - | </code> | + | |
| - | + | ||
| - | or written differently | + | |
| - | + | ||
| - | <code> | + | |
| - | Sum[Product[1 - (k - j)/n, {j, 1, k}] z^k/k!, {k, 0, n}] // Simplify | + | |
| - | + | ||
| - | Sum[Product[-(n - j), {j, 1, k}] (1/(k + 1)) z^k/k!, {k, 0, n}] // Simplify | + | |
| - | </code> | + | |
| === References === | === References === | ||
