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There is a well-known estimate for the sum of all binomial coefficients $\binom{n}{k}$ satisfying $k \leq \alpha n$ for some $\alpha$ satisfying $0 < \alpha \leq 1/2$:

$$ \sum_{k=0}^{\alpha n}\binom{n}{k} = 2^{(H(\alpha) + o(1))n}$$

where $H(\alpha) = -\alpha\log_2(\alpha) - (1-\alpha)\log_2(1-\alpha)$ is the binary entropy function.

My question is whether there exists a similar estimate when we weight the $k$-th binomial coefficient by $\lambda^k$ for some $\lambda > 0$. That is, I would like to estimate the following sum:

$$ \sum_{k=0}^{\alpha n} \binom{n}{k} \lambda^k $$

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Have you tried replacing 2 by (1+lambda)? Alternatively, for large lambda and alpha far enough away from 1/2, some nice estimates occur by using geometric series approximations. Gerhard "Ask Me About System Design" Paseman, 2012.004.11 – Gerhard Paseman Apr 11 '12 at 10:12
There are many estimates here. Is there context as to what kind of estimate you are looking for? – Daniel Parry Apr 11 '12 at 18:34
up vote 6 down vote accepted

Expanding on the previous answers. I'm taking $\lambda$ and $\alpha$ to be constants which do not vary as $n\to\infty$.

If $α<λ/(λ+1)$ then the sum is within a constant of the last term. In fact the largest terms are approximately in geometric progression so you can get it quite accurately by computing the ratio.

If $α>λ/(λ+1)$, almost all of the complete binomial expansion is present, so the sum equals $(1+o(1))(1+λ)^n$.

If $α≈p$, then Russell's normal approximation will be good. (This needs some work to clarify whether the geometric approximation of the lower tail is good right up to the point where the normal approximation begins to be good. I think it is.)

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The last assertion is actually not true, as stated: say, if $\lambda=1$ and $\alpha$ is just a tiny bit larger than $1/2$, then the sum is not equal to $(1+o(1))2^n$ (it is half of this quantity). – Seva Apr 11 '12 at 14:02
Perhaps you are not taking $\alpha$ to be constant as $n\to\infty$. Apart from the scaling factor of $2^n$, it is a binomial distribution $Bin(n,1/2)$, which has a standard deviation of $\sqrt{n}/2$. So if $\alpha=1/2+\epsilon$ for fixed $\epsilon>0$, the upper limit of your sum is $2\epsilon\sqrt{n}$ standard deviations above the mean, so it cuts off a negligible portion of the sum from 0 to $n$. To asymptotically get half of $2^n$ you need $\alpha=1/2+o(n^{-1/2})$, which is not constant except for $\alpha=1/2$. – Brendan McKay Apr 12 '12 at 2:08
What happens when $\alpha=c/n$ for $c$ a constant? Can one bound the sum in terms of $c$, $\lambda$ and $n$? – Bullmoose Nov 22 '13 at 4:33
@Bullmoose: In that case the sum is strongly dominated by its largest term. – Brendan McKay Nov 22 '13 at 4:50
I posted a question related to this here:… The question is the origin of my query. – Bullmoose Nov 22 '13 at 5:57

Your sum can also be thought of as the first $\alpha n$ terms in a binomial distribution with probability of success $p=1-\frac1{\lambda+1}$. So, it is closely approximated by a normal distribution with mean $np$ and standard deviation $\sqrt{np(1-p)}$, i.e., $$\sum_{k=0}^{\alpha n} \binom nk \lambda^k\approx (1-p)^{-n}\Phi\left((\alpha-p)\sqrt{\frac n{p(1-p)}}\right),$$ where $\Phi$ is the cumulative standard normal distribution.

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This will only be a good estimate when the upper limit is close to the centre of the normal approximation, i.e., when $\alpha\approx p$. – Brendan McKay Apr 11 '12 at 11:39

Sums of this sort are estimated by their largest summand, and the resulting estimate will depend on the relation between $\alpha$ and $\lambda$.

The ratio of the $(k+1)$th and the $k$th terms of the untruncated sum is $$ \lambda \frac{n-k}{k+1}, $$ showing that the sequence of summands is unimodal, with the maximum value attained for $k$ about $\frac{\lambda}{\lambda+1}\,n$. Consequently, if $\alpha<\lambda/(\lambda+1)$, then your sum is between $\binom n{\lfloor\alpha n\rfloor}\lambda^{\lfloor\alpha n\rfloor}$ and $n\binom n{\lfloor\alpha n\rfloor}\lambda^{\lfloor\alpha n\rfloor}$, which is $2^{H((\alpha)+\alpha\log_2\lambda+o(1))n}$; similarly, if $\alpha>\lambda/(\lambda+1)$, then the sum is $2^{(H(\lambda/(\lambda+1))+(\lambda/(\lambda+1))\log_2\lambda+o(1))n}=(\lambda+1)^{1+o(1)}$. (Both estimates assume that $\alpha$ and $\lambda$ are fixed, and $n\to\infty$.)

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