Can you simplify (or approximate) $\sum_{n=0}^{N-1} \binom{N-1}n \frac{(-1)^n}{n+1} e^{-\frac{n}{2(n+1)}\lambda}$? Let $\binom x y$ be the binomial coefficient. I am trying to get a better understanding of the sum
$$
f(N,\lambda)=\sum_{n=0}^{N-1}\binom{N-1}n\frac{(-1)^n}{n+1} e^{-\frac{n}{2(n+1)}\lambda}
$$
as a function of $\lambda$ and $N$ without having to plot it for all possible values.
Do you know of a way to simplify it or approximate it? Although I'd rather obtain an approximation for positive $\lambda$ and $N$, an asymptotic approximation for $\lambda\rightarrow\infty$ or $N\rightarrow\infty$, or even an upper-bound would be helpful too.
Here is an equivalent expression:
$$
f(N,\lambda)=\frac{1}{N}\sum_{n=1}^{N}\binom N n(-1)^{n-1} e^{-\frac{n-1}{2n}\lambda}.
$$
In my opinion, this sum could be easily simplified if the exponential term would have an argument that was not a rational function $\frac{n}{n+1}$ or $\frac{n-1}{n}$. I have not found a way in the literature to deal with it.
 A: The factor $\exp(-n\lambda/(2n+2))$ probably precludes significant
simplification of $f(N,\lambda)$, but one can still construct
an integral representation that shows $f(N,\lambda) > 0$ for all
positive $N$ and $\lambda$, and makes it possible to estimate the sum.
The idea is to find, given $\lambda$, a representation of
$\exp(-n\lambda/(2n+2)) \, / \, (n+1)$ as a linear combination of
exponentials $\exp(-cn)$ for which the sum can be evaluated
in closed form as a binomial expansion.  In other words,
we want to write $\exp(-n\lambda/(2n+2)) \, / \, (n+1)$ as a Laplace transform.
We use the definite integral formula
$$
\int_0^\infty I_0(\sqrt x) e^{-ax} \, dx = \frac1a \exp \frac1{4a},
$$
where $I_0$ is the modified Bessel function
$I_0(z) = \sum_{m=0}^\infty (z/2)^{2m} / m!^2$.  Writing
$$
e^{-\frac{n}{2(n+1)}\lambda} = e^{-\lambda/2} e^{\frac\lambda{2(n+1)}},
$$
we soon find
$$
f(N,\lambda) = \frac{e^{-\lambda/2}}{2\lambda}
  \int_0^\infty I_0(\sqrt x) \left[
    \sum_{n=0}^{N-1} (-1)^n {N-1 \choose n} e^{-\frac{n+1}{2\lambda}x}
  \right] \, dx
$$
in which the bracketed sum is $\exp(-x/2\lambda)$ times a binomial expansion:
$$
f(N,\lambda) = \frac{e^{-\lambda/2}}{2\lambda}
  \int_0^\infty I_0(\sqrt x) \, e^{-\frac{x}{2\lambda}}
    \left(1 - e^{-\frac{x}{2\lambda}} \right)^{N-1}
  \, dx \, .
$$
The integrand is nonnegative, so $f(N,\lambda) > 0$ as claimed.
Moreover, its size can be estimated for any given $N,\lambda$
by determining at what $x$ the integrand is maximized and how rapidly
it decays from this maximum.
A: for $|\lambda|\ll 1$ you might expand as
$$Nf(N,\lambda)=1+\tfrac{1}{2}\lambda\,(H_N-1)+{\rm order}\,(\lambda^2)$$
with $H_N$ the harmonic number.
For $\lambda$ large a reasonable approximation is
$$f(N,\lambda)\approx 1-\tfrac{1}{2}(N-1)e^{-\lambda/4},\;{\rm for}\;\lambda\,{\gtrsim}\, 1,N$$
