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So here is the problem, and what I've worked through.. Although I am not sure if I am doing it correctly, as it is very confusing to me. The goal is to find the interval and radius of convergence of a power series.

$\sum _{k=1}^{\infty }\:\frac{1}{k}\left(x-1\right)^{2k+1}$

$lim_{k→∞}⁡|\frac{a_{k+1}}a|=lim_{k→∞}|\frac{1}{k}(x-1)^{(2(k+1)+1)}| |\frac{k}{1}*\frac{1}{(x-1)^{2k+1}}|=lim_{k→∞}⁡|\frac{(x-1)^{2}*k}{k+1}|$

$lim_{k→∞}⁡|\frac{\frac{(x-1)^2*k}{k+1}}{\frac{k}{k}+\frac{1}{k}}|=lim_{k→∞}|\frac{(x-1)^2}{1}|=lim_{k→∞}|(x-1)^2|<1$

$-1<(x-1)^2<1$

$0<x<2$

For $x=0$

$\sum_{k=1}^\infty\frac{1}{k}(0-1)^{2k+1}=\sum_{k=1}^\infty(-1)^{2k+1}*\frac{1}{k}\ = lim_{k\to\infty}|\frac{1}{k}|=0$

For $x=2$

$\sum_{k=1}^\infty\frac{1}{k}(2-1)^{2k+1}=\sum_{k=1}^\infty(1)^{2k+1}*\frac{1}{k}= lim_{k\to\infty}|\frac{a_{k+1}}{a_{k}}|=lim_{k\to\infty}\frac{\frac{1}{k+1}}{\frac{1}{k}}=lim_{k\to\infty}\frac{k}{k+1}\frac{\frac{1}{k}}{\frac{1}{k}}=1$

Therefore, the radius of convergence $r=\frac{2-0}{2}=1$, and the interval of convergence is $(0,2)$

I feel very confused when trying to work through these.

xerxes2985
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    Please clean this up a bit. To name some problems, you have written equalities that aren't true (limits on one side, but not on the other), messed up indices immediately and written implications i dont understand. If $x=0$ then that is indeed the (convergent) sum, then you write an arrow and state a true fact about a limit. What is your claim. The second $x=0$ should also be an $x=2$. – operatorerror Mar 18 '18 at 02:49
  • @qbert I've just umm.. well tried to clean it up a bit... I don't quite get the aspect of what the $\to$ is used for... I just think of it like ... well this becomes this, and etc.. – xerxes2985 Mar 18 '18 at 03:00
  • it's usually an implication (actually, $\Rightarrow$ is), but even with this interpretation, what you wrote didn't make sense. Even right now, that sum is not equal to the limit as $k\to \infty$ of $|1/k|$? – operatorerror Mar 18 '18 at 03:39

2 Answers2

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Your confusion is at the boundary point $x=0$.

$$\sum _{k=1}^{\infty }\:\frac{1}{k}\left(x-1\right)^{2k+1}$$ Note that for x=0 you have

$$\sum _{k=1}^{\infty }\:\frac{1}{k}\left(-1\right)^{2k+1}$$ which is an alternating harmonic series and it is convergent.

Thus the interval of convergence is $[0,2)$

The rest of your work is flawless.

Mohammad Riazi-Kermani
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Here is another quick way to solve this problem using Cauchy's Criterion:

$$ \begin{align} \sqrt[k]{\left\lvert a_k\right\rvert} &=\sqrt[k]{\left\lvert\frac{1}{k}(x-1)^{2k+1}\right\rvert} =\frac{1}{\sqrt[k]{k}}\left\lvert x-1\right\rvert^{\frac{2k+1}{k}} =\frac{1}{e^{\frac{ln(k)}{k}}}\left\lvert x-1\right\rvert^{2}\left\lvert x-1\right\rvert^{\frac{1}{k}}\\ &\to\frac{1}{e^{0}}\left\lvert x-1\right\rvert^{2}\left\lvert x-1\right\rvert^{0} =\left\lvert x-1\right\rvert^2<1\Rightarrow \left\lvert x-1\right\rvert<1\Leftrightarrow 0<x<2 \end{align} $$

Then, based on the above, the radius of convergence is $1$ and the interval is $[0, 2)$, because:

Angel Politis
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