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This was a thought question assigned to our calc II class, and I wasn't sure how to approach it.

Give an example of a power series whose interval of convergence is $(0, \frac{4}{3}]$. Show that it has this interval of convergence.

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  • $\begingroup$ Try $\sum \frac{(x-2/3)^n}{n}$. It is by the way not quite right, but the fix should not be hard. $\endgroup$ – André Nicolas May 2 '15 at 23:58
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I'll start with something I know and see if I can get to what you want.

First of all, you want a series that converges on a half-open interval. The simplest one I can think of is $\sum_{n=1}^{\infty} \frac{x^n}{n}$ which converges for $-1 \le x < 1$.

To make this converge at the upper limit, change the sign of $x$ so the series becomes $\sum_{n=1}^{\infty} \frac{(-x)^n}{n}$ which converges for $-1 < x \le 1$.

The interval of convergence is $2$ ($=1-(-1)$). To make it $4/3$, multiply $x$ by $\frac{2}{4/3} =\frac{3}{2} $. The series then becomes $\sum_{n=1}^{\infty} \frac{(-3x/2)^n}{n}$ which converges for $-\frac23 < x \le \frac23$.

Finally, you want the interval of convergence to start at $0$ instead of $-2/3$. To do this, subtract $2/3$ from $x$, making the series $\sum_{n=1}^{\infty} \frac{(-3(x-2/3)/2)^n}{n} =\sum_{n=1}^{\infty} \frac{(-3x/2+1)^n}{n} $ which converges for $0 < x \le \frac43$.

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Edit (the incorrect answer previously given was due to a careless reading):

Take the Taylor series of $ln(x+1)$, which has an interval of convergence of $(-1, 1]$. Then substitute $x$ with $\frac 32 x - 1$ in the series.

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  • $\begingroup$ So, you wind up with pretty much the same answer as marty cohen's. $\endgroup$ – Gerry Myerson May 3 '15 at 3:30

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