Ratio test for $\sum_{n=1}^\infty \frac{\log n}{n^2}z^n$

Im trying to find the radius of convergence for $\sum_{n=1}^\infty \frac{\log n}{n^2}z^n$. Applying the ratio test $\frac{C_{n+1}}{C_n}$, I simplified $\frac{\log n}{n^2}$ to $\frac{\log(n+1)}{\log n}\cdot\frac{n^2}{(n+1)^2}$.

Would I be correct in saying the radius of convergence is 1, by applying the squeeze theorem: $\frac{n^2}{(n+1)^2} < \frac{\log(n+1)}{\log n}\cdot\frac{n^2}{(n+1)^2} < \frac{\log(n+1)}{\log n}$ because the limit as $n$ tends to infinity of both $\frac{\log(n+1)}{\log n}$ and $\frac{n^2}{(n+1)^2}$ is 1?

Thanks.

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You're correct in saying that that limit is $1$, but that's not at all the same thing as saying the radius of convergence is $1$.

You need to look at $$\frac{(\log (n+1)) |z^{n+1}|/(n+1)^2}{(\log n)|z^n|/n^2} = \frac{\log(n+1)}{\log n}\cdot\frac{n^2}{(n+1)^2}\cdot |z| \to |z| \text{ as } n\to\infty.$$ This limit is less than $1$ if $|z|<1$ and greater than $1$ if $|z|>1$. Therefore the radius of convergence is $1$.

Whether it converges when $|z|=1$ may be a more difficult question.

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Your argument is correct. There was a verbal slip, in that it is not true that you simplified $\frac{\log n}{n^2}$ to $\frac{\log(n+1)}{\log n}\cdot \frac{n^2}{(n+1)^2}$. What you did was to simplify something else, which is a ratio of successive coefficients.

Applying the Squeeze Theorem, though well done, was unnecessary. You are trying to find $$\lim_{n\to\infty}\frac{\log(n+1)}{\log n}\cdot \frac{n^2}{(n+1)^2}.$$ There is a general result that says that if $\lim_{n\to \infty}a_n=a$ and $\lim_{n\to \infty}b_n=b$, where $a$ and $b$ are real numbers, then $$\lim_{n\to \infty}a_nb_n=ab$$ (the limit of a product is the product of the limits). That's all you need.

Comment: Getting practice in using the Squeeze Theorem is not a bad idea. However, inequalities of the type you found are not always so easy to get. That's why certain basic tools, about the limit of a sum, and the limit of a product, are so useful. Another very useful result if that if the sequence $(a_n)$ has limit $a$, where $a$ is a real number, and $f$ is a continuous function, then the sequence $(f(a_n))$ has limit $f(a)$.

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The argument for the statement that that particular limit is $1$ is correct, but the claimed conclusion was that the radius of convergence is $1$. –  Michael Hardy Nov 10 '11 at 0:08
@Michael Hardy: In a post earlier today, the OP used explicitly the result that the radius of convergence is, sort of, the reciprocal of the limit. I would prefer that he calculate explicitly with the $z$ in place, like you did, instead of using a canned formula. –  André Nicolas Nov 10 '11 at 0:17

What you write is true, but you don't need to use the squeeze theorem for most problems like this (even though you used it correctly).

You can just use properties of limits, splitting the limit up as a product of limits.

Since $\displaystyle \lim_{n\rightarrow \infty} \frac{\ln (n+1)}{\ln (n)} = 1$ and $\displaystyle \lim_{n\rightarrow \infty} \frac{n^2}{(n+1)^2}=1$, $$\lim_{n\rightarrow \infty} \left( \frac{\ln (n+1)}{\ln (n)} \cdot \frac{n^2}{(n+1)^2} \right)=1 \cdot 1 = 1.$$

I'm not sure what reasoning you're using to decide that the radius of convergence is 1. It is true that if you have $\displaystyle \sum_{n=1}^\infty c_n z^n$ and $\displaystyle \lim_{n \rightarrow \infty} \frac{|c_{n+1}|}{|c_n|}=r$, then the radius of convergence of the power series is $\frac{1}{r}$.

So the radius of convergence is $\frac{1}{1}=1$. For a more direct way to work out the radius of convergence, you should look at Michael Hardy's answer.

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