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Anthony Quas
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Do you know how good you need it? Provided $k < n/3$ say, a reasonable bound (correct to within a fixed multiplicative factor of 2) is obtained by taking the last term $\binom {n}{k}$ (you see this because you can compute the ratio of each term to the prior term and bound it above by 1/2. Now you can estimate the sum as a geometric series.)

For $\sum_{j < k} \binom{n}{j}a^j(1-a)^{n-j}$, bounding by the last term also works quite well as long as $k$ is a good bit smaller than $an$.

Do you know how good you need it? Provided $k < n/3$ say, a reasonable bound (correct to within a fixed multiplicative factor) is obtained by taking the last term $\binom {n}{k}$ (you see this because you can compute the ratio of each term to the prior term and bound it above by 1/2. Now you can estimate the sum as a geometric series.)

For $\sum_{j < k} \binom{n}{j}a^j(1-a)^{n-j}$, bounding by the last term also works quite well as long as $k$ is a good bit smaller than $an$.

Do you know how good you need it? Provided $k < n/3$ say, a reasonable bound (correct to within a multiplicative factor of 2) is obtained by taking the last term $\binom {n}{k}$ (you see this because you can compute the ratio of each term to the prior term and bound it above by 1/2. Now you can estimate the sum as a geometric series.)

For $\sum_{j < k} \binom{n}{j}a^j(1-a)^{n-j}$, bounding by the last term also works quite well as long as $k$ is a good bit smaller than $an$.

Source Link
Anthony Quas
  • 23.2k
  • 5
  • 63
  • 98

Do you know how good you need it? Provided $k < n/3$ say, a reasonable bound (correct to within a fixed multiplicative factor) is obtained by taking the last term $\binom {n}{k}$ (you see this because you can compute the ratio of each term to the prior term and bound it above by 1/2. Now you can estimate the sum as a geometric series.)

For $\sum_{j < k} \binom{n}{j}a^j(1-a)^{n-j}$, bounding by the last term also works quite well as long as $k$ is a good bit smaller than $an$.