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given the series

$$ f(x)= \sum_{n=0}^{\infty}\frac{(2x)^{n}}{n!}(-1)^{n}\frac{d^{n}}{dx^{n}}(\frac{1}{x-1})$$

how could i evaluate this for every x different from x=1 ?? thanks any hints?

or if possile the analogue series $$ g(m)=\sum_{n=0}^{\infty} \frac{1}{n!}\int_{0}^{\infty}dxln^{n}(x)x^{-m}$$

Jose Garcia
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2 Answers2

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Hint: For every $n\geqslant0$, $\displaystyle\frac{\mathrm d^n}{\mathrm dx^n}\frac1{1-x}=\frac{\underline{\qquad}}{(1-x)^{n+1}}$ hence $$f(x)=\displaystyle\frac1{x-1}\frac1{1-\frac{2x}{x-1}}=-\frac1{1+x}.$$

Did
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For f(x) you will find if you expand the second bit, it gets significantly less complicated.

Write the first few terms of $ \frac{d^{n}}{dx^{n}}(\frac{1}{x-1})$ and you should be able to guess $ \frac{d^{n}}{dx^{n}}(\frac{1}{x-1}) = \frac{(-1)^n(n!)}{(x-1)^{n+1}}$ and then prove this by induction.

Then $$ f(x) =\sum_{n=0}^\infty \frac{(2x)^n}{(x+1)^{n+1}}$$

Hopefully you can take it from there?

  • Nice to spoil the question by providing a full answer after someone took pain to leave something to do to the OP. If you wanted to write a full answer, you could at least have chosen the second part of the question to do so, don't you think? – Did Jan 17 '14 at 21:52
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    ok thanks :D i didn't know how i cuold not get this solution by myself.. – Jose Garcia Jan 17 '14 at 22:37