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

Find the Taylor series for this function around $x=1$.

I don't understand this, isn't that summation already the Taylor polynomial of $f(x)$? If I follow the procedure of Taylor's theorem and differentiate I get the exact same result again.

Also it's a power series and by Taylor's theorem the coefficients $a_k$ should be equal to $f^{(k)}(1)$, which, non-surprisingly, they are.

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    Perhaps they want you to evaluate the sum? – 2017-01-25
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    Find the $f(x)$ it represents? – 2017-01-25
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    Perhaps. In closed form or something. That might be what they really want, though it seems unclear. – 2017-01-25
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    Any idea on how to do that though? I suppose it's convergence radius is 1 but I'd have to check that. These exercises are really hard, they basically explain you nothing but Taylor's theorem and I'm expected to be able to reverse that process and all.. Don't really understand what's going on here. – 2017-01-25
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    Radius of convergence should be $2$. And, if I may, I'd point to the polylogarithm. – 2017-01-25

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Taking upon the semi-agreed question as per the comments, your function is:

$$f(x)=\frac{2\ln\left(\frac12(3-x)\right)}{1-x}$$

As per the Taylor series of the natural logarithm:

$$\ln(1-x)=\sum_{k=1}^\infty\frac{x^k}k$$

and the given Taylor series has radius of convergence $R=2$ by the ratio test.

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    I'm sorry but I don't understand that. I get that the standardpolynomial can be applied to $ln(2-x)$ by using $x=x-1$, but how does the rest fit in? And how to get the sum to start at $k=0$? – 2017-01-25
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    @QuestionMaker $$\sum_{k=0}^\infty a_k=\sum_{k=1}^\infty a_{k-1}$$The rest should fit in quite nicely :-) – 2017-01-25
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    But wouldn't you get: $$\ln(1-(x-1))=\ln(2-x)=\sum_{k=0}^\infty\frac{(x-1)^{k+1}}{k+1}$$ – 2017-01-25
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    @QuestionMaker Oh, right, I forgot the $1/2$. Thanks for the catch! XD – 2017-01-25
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    Ok but how would you use the standard polynomial of the natural logarithm for something like $(3-x)/2$ – 2017-01-25
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    @QuestionMaker Usually, I'd rewrite it as $\ln(1-\dots)$ and then apply Taylor expansion of $\ln$. – 2017-01-25
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    $$\ln(1-(x/2-1/2))=-\sum_{k=0}^\infty\frac{(x/2-1/2)^{k+1}}{k+1}$$ – 2017-01-25
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    @QuestionMaker Mhm. Now we have a $(x/2-1/2)^{-1}$ out in the front, so it cancels a bit, and what do you get when the dust clears? – 2017-01-25
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    What do you mean by out in the front? Multiplying this summation by $2/(1-x)$ doesn't solve much for me.. – 2017-01-25
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    Oh the multiplication gives $$-\sum_{k=0}^\infty\frac{(x/2-1/2)^{k}}{k+1}$$ – 2017-01-25
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    Then factor the $1/2$ out and you get your series. :-) – 2017-01-25
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    $$-\sum_{k=0}^\infty\frac{(x/2-1/2)^{k}}{k+1}=-\sum_{k=0}^\infty\frac{(x-1)^{k}}{2^k(k+1)}$$ But I should still get rid of the minus which is in the polynomial for the natural logarithm (look it up, I think you forgot it in your main post) – 2017-01-25
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    @QuestionMaker Nah. You just forgot that $2/(1-x)$ is equal to $-(x/2-1/2)^{-1}$ due to the order I wrote it. And then the negatives all go away :D – 2017-01-25
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    Oops stupid of me. Well thanks a great lot for helping me out. Shed some light on how to use those standard series too. – 2017-01-25
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    :D Glad to help. (and for those standard series, I usually just ask Google for "Taylor series" and hit "images") – 2017-01-25