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How can I simplify this summation $$\sum_{i=1}^\infty \left[1-\sum_{n=0}^{i-1}(-1)^n \frac{a^n}{n!} \left. \frac{d^n}{dt^n} f(t)\right|_{t=a} \right] $$

if $f(t)$ is equal to $\left(\dfrac{b}{t+b}\right)^2$

I thought about using taylor series where $f(x)=f(c)+f'(c)(x-c)+f''(b)(x-c)^2/2!+\cdots.$ Now taking $x=c-a$ then $f(c-a)=f(c)-a.f'(c)+f''(c)a^2/2!-\cdots+.$

An update: from the cauchy integeal $$f(z)=\sum_{n=0}^\infty (z-z_0)^n \frac{f^{(n)}(z_0)}{n!}$$ then taking $z_0=a$ and $z=0$ $$f(0)=\sum_{n=0}^\infty (-1)^n a^n \frac{f^{(n)}(a)}{n!}$$

ahmed ayman
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  • any own thoughts? – tired Nov 12 '15 at 19:32
  • It seems like it is in the form of tylor expansion. $f(a)-af'(a)+.....+ \frac {a^n}{n!}/f''(a)$.... Some technical papers uses a similar equation and the results in a finite number depending on $a$. – ahmed ayman Nov 12 '15 at 19:36

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With help of a nice professor, this summation is approximated to two geometric series. $$f'(a)= \dfrac{-2b^2}{(a+b)^2}$$ $$f''(a)= \dfrac{6b^2}{(a+b)^3}$$ $$f^{(n)}(a)= \dfrac{(-1)^n(n+1)!b^2}{(a+b)^{n+2}}$$ sub. in the original equation $$\sum_{i=1}^\infty [1-\sum_{n=0}^{i-1} \frac{a^nb^2(n+1)}{(a+b)^{n+2}}]$$ $$\sum_{i=1}^\infty [1-\left (\dfrac{b}{a+b}\right)^2\sum_{n=0}^{i-1} (n+1)\left (\dfrac{a}{a+b}\right)^n]$$ $$\sum_{i=1}^\infty [1-\left (\dfrac{b}{a+b}\right)^2[\sum_{n=0}^{i-1} \left (\dfrac{a}{a+b}\right)^n+\sum_{n=0}^{i-1} n\left (\dfrac{a}{a+b}\right)^n]]$$ The first term is a geometric series and the second term is first derivative of a geometric series.

The final result will be $$\left(\dfrac{b+2a}{b}\right)$$

ahmed ayman
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