Evaluating a sum by Abel's Thm

[benorin]

I wish to evaluate the sum

$$\sum_{k=n}^{\infty} \left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^k}{k+1}$$

I think I can use Abel's theorem... just read: for |z|<1, we have

$$\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = z^n\frac{1}{(1+z)^{n+1}} = (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \frac{1}{1+z}\right)$$
$$= (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \sum_{k=0}^{\infty}(-1)^{k}z^{k} \right) = (-1)^n\frac{z^n}{n!}\sum_{k=n}^{\infty}(-1)^{k}\frac{k!}{(k-n)!}z^{k-n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k}$$

here obtained is the identity

$$\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k}$$​

valid for |z|<1. Suppose that 0<x<1 and multiply the above identity by $$(-1)^n$$ and then integrate over (0,x) to arrive at

$$\int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \int_{z=0}^{x} \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k}dz = \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) \frac{x^{k+1}}{k+1}$$​

to the latter sum apply Abel's Theorem to determine that

$$\lim_{x\rightarrow 1^-} \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) \frac{x^{k+1}}{k+1} = \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1}$$​

so that, by the previous identity, the value of the sum is given by

$$\boxed{ \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} = \lim_{x\rightarrow 1^-} \int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} }$$​

Okay, so www.integrals.com[/url] (also, [PLAIN]http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced#reply ) gives the integral in terms of the hypergeometric function ₂F₁, viz.

$$\lim_{x\rightarrow 1^-} \int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \lim_{x\rightarrow 1^-} (-1)^k\frac{x^{k+1}}{k+1} _2F_1 (k+1,k+1;k+2;-x) = \frac{(-1)^k}{k+1} _2F_1 (k+1,k+1;k+2;-1)$$​

so here I need assistance, the sum above is to be again summed over, as in

$$\zeta ^{\prime} (0) =\sum_{n=0}^{\infty} (n+1)\left[ \log(n+1) -1\right] \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1}$$​

where $$\zeta (s)$$ denotes Riemann's zeta function, and it should sum to $$-\log \sqrt{2\pi}$$: and thoughts as to how to proceed from here?

 

[benorin]

That integral has promise...

That integral has promise... so put

$$I_{n}=\int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z}$$

my prof said to try integration by parts (thanks Mihai,) pick, say

$$u=z^n\Rightarrow du=nz^{n-1}dz \mbox{ and } dv=(-1)^{n}(1+z)^{-n-1}dz\Rightarrow v=\frac{(-1)^{n+1}}{n}(1+z)^{-n}$$​

which gives

$$I_{n} = \int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \left[ -\frac{1}{n} \left( \frac{-z}{1+z} \right) ^{n} \right] _{0}^{1} - \int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n-1}\frac{dz}{1+z} = \frac{(-1)^{n+1}}{n2^n} - I_{n-1} \,$$

a recurrence relation: I'll try using the generating function technique. To be continued...

 

[benorin]

generating functions are sweet

About that generating function... we will need the value $$I_0=\log 2$$ (use the series or the integral, both are easy.) From the recurrence relation

$$I_{n}= \frac{(-1)^{n+1}}{n2^n} - I_{n-1} \,$$​

multiply by $$z^n$$ and sum over $$n\geq 1$$ to get

$$G(z):=\sum_{n=1}^{\infty} I_{n}z^{n} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n2^n}z^{n} - \sum_{n=1}^{\infty}I_{n-1}z^{n} = \log \left( 1+ \frac{z}{2}\right) -zI_0-z \sum_{n=1}^{\infty}I_{n}z^{n}$$​

if we put $$G(z) = \sum_{n=1}^{\infty} I_{n}z^{n}$$ for the generating function, substitute in the value of $$I_0=\log 2$$ we have

$$G(z)= \log \left( 1+ \frac{z}{2}\right) -z\log 2 -zG(z)$$​

and hence

$$G(z)= \frac{\log \left( 1+ \frac{z}{2}\right) -z\log 2}{1+z} = \left( \sum_{n=0}^{\infty} (-1)^{n}z^n\right) \left( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1}2^{-n}z^{n}\right) \frac{z}{2} - z\log 2 \sum_{n=0}^{\infty} (-1)^{n}z^n$$
$$= \frac{z}{2}\sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{(-1)^{k}}{k+1}2^{-k}(-1)^{n-k}\right) z^{n} - \log 2 \sum_{n=1}^{\infty} (-1)^{n-1}z^{n}$$
$$= \sum_{n=1}^{\infty} (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} \right) z^{n} - \log 2 \sum_{n=1}^{\infty} (-1)^{n}z^{n} = \sum_{n=1}^{\infty} (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right) z^{n}$$​

so we can now equate coefficients of $$z^{n}$$ in

$$\sum_{n=1}^{\infty} I_{n}z^{n}= \sum_{n=1}^{\infty} (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right) z^{n}$$​

to arrive at

$$I_{n}=\int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right)$$​

which (finally) gives the value of the desired sum

$$\boxed{ \sum_{k=n}^{\infty} \left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^k}{k+1} = \left\{\begin{array}{cc}\log 2,&\mbox{ if }<br /> n=0\\(-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right), & \mbox{ if } n\geq 1\end{array}\right. \, }$$​