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Radius of convergence is for power series but how does one go about computing the radius of convergence of the infinite sum $\sum_{k=1}^{\infty} \frac{k}{k+1}\left(\frac{2x+1}{x}\right)^k\ ?$

Can you find the radius $R$ directly by the standard $1/\limsup$ formula (or the Ratio Test) or do you have to make some kind of substitution to get it into the right form? I got the domain of convergence $(-1, -1/3)$ when I used the Ratio Test.

If you do need a substitution, give me a hint of how to go about that. If I am okay computing $R$ correctly with Ratio or Root Test, give me a hint why that's acceptable (even though $(2x+1)/x$ isn't a term of a polynomial).

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You can use the standard formulas, since for fixed $x$, you get convergence or divergence of the series. Using the ratio test, one gets that $ \left| \frac{a_{k+1}}{a_k} \right| = \frac{(k+1)^2}{k(k+2)} \left| \frac{2x+1}{x} \right| \to L < 1 $ if and only if $|2x+1| < |x|$ with $x \neq 0$ (after computing). This gives you the interval $(-1,-1/3)$ indeed.

Hope that helps,

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    One thing though ; your nickname is kinda fancy for such a welcoming website. Heh! Just saying.2012-03-07
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Yes, the Ratio Test will work. After forming and simplifying $\left|\frac{x_{n+1}}{x_n}\right|$, it will (typically) involve the variable $x$. Take the limit as $n\to\infty$ and then figure out what x's will make the limit $L<1$. Convergence for the case when $L=1$ has to be decided some other way.

The idea of radius of convergence applies to any infinite series involving a variable, $x$ for example.

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If you set $y=\frac{2x+1}{x}$, your series rewrites: $\tag{1} \sum_{k=1}^\infty \frac{k}{k+1}\ y^k$ which is a power series.

The radius of convergence of (1) equals $1$, and (1) converges also for $y=-1$; therefore the convergence set of (1) is $[-1,1[$, i.e.: $\tag{2} -1\leq y<1$

Finally, you can return to your original variable setting $y=\frac{2x+1}{x}$ in (2); solving (2) w.r.t. $x$ gives the convergence set of your original series.

In fact your original series converges if and only if $x$ solves: $\begin{cases} \frac{2x+1}{x}\geq -1 \\ \frac{2x+1}{x}<1 \end{cases} \quad \Leftrightarrow \quad \begin{cases} \frac{3x+1}{x}\geq 0 \\ \frac{x+1}{x}<0 \end{cases} $ hence iff $x\in ]-1 ,-1/3]$.