# Thread: What power should I use for a root test over a series?

1. ## What power should I use for a root test over a series?

To test for convergence of a series, I would usually take the limit of the series with root test's power 1/n.

But in some cases, I was told that I could use the power of the coefficient of a power series. For instance, in this series ${\textstyle \underset{p=2}{\sum}\frac{(4)^{2p}}{9^{p}p^{5/4}}(\frac{5}{4}-x)^{2p}}$, its coefficient has a power of 2p. I could just take the limit of the coefficient, $\underset{p\rightarrow\infty}{\lim}\frac{1}{\left[\frac{(4)^{2p}}{9^{p}p^{5/4}}\right]^{1/2p}}=\frac{3}{4}$. So 3/4 is the radius of convergence, which is correct. If I took the limit of the root test with power 1/n, the answer would be wrong.

But then in another series like this one: $\underset{n=2}{\sum}\frac{n^{n^{2}}e^{3n}}{(2+n)^{ n^{2}}2^{2n}}4^{2n}(x+\frac{5}{4})^{2n}$, when I take the limit of its coefficient with its power 2n, $\underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\cdot e^{1/2}}$, the answer for the radius of convergence is wrong. The correct answer is 1/(4e). Apparently, I will only get the 1/(4e) as the answer if I took 1/n in my root test instead of 1/2n.

In this case, how would I know when I should take 1/n or 1/2n or even 1/(2n+1) in my root test for a series?
Thanks!

2. Originally Posted by xEnOn
To test for convergence of a series, I would usually take the limit of the series with root test's power 1/n.

But in some cases, I was told that I could use the power of the coefficient of a power series. For instance, in this series ${\textstyle \underset{p=2}{\sum}\frac{(4)^{2p}}{9^{p}p^{5/4}}(\frac{5}{4}-x)^{2p}}$, its coefficient has a power of 2p. I could just take the limit of the coefficient, $\underset{p\rightarrow\infty}{\lim}\frac{1}{\left[\frac{(4)^{2p}}{9^{p}p^{5/4}}\right]^{1/2p}}=\frac{3}{4}$. So 3/4 is the radius of convergence, which is correct. If I took the limit of the root test with power 1/n, the answer would be wrong.

But then in another series like this one: $\underset{n=2}{\sum}\frac{n^{n^{2}}e^{3n}}{(2+n)^{ n^{2}}2^{2n}}4^{2n}(x+\frac{5}{4})^{2n}$, when I take the limit of its coefficient with its power 2n, $\underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\cdot e^{1/2}}$, the answer for the radius of convergence is wrong. The correct answer is 1/(4e). Apparently, I will only get the 1/(4e) as the answer if I took 1/n in my root test instead of 1/2n.

In this case, how would I know when I should take 1/n or 1/2n or even 1/(2n+1) in my root test for a series?
Thanks!

It is ALL the same! In the first example you use p since that is the running index in the power

series, and in the second you should use n because that is the index....and if there's an example of a power series

with running index equal to Gua, then you must use the Gua-root test...

Tonio

3. The running index variable is p. But in the first example, I used 1/(2p) as my root test to get the correct answer because I was told that the power of the co-efficient of the series is 2p and I need to use 1/(2p).

But in the second example, the co-efficient also has 2n as its power but when I use 1/(2n) for the root test, it gives me a wrong answer.

So it doesn't look quite the same, isn't it?

4. Originally Posted by xEnOn
The running index variable is p. But in the first example, I used 1/(2p) as my root test to get the correct answer because I was told that the power of the co-efficient of the series is 2p and I need to use 1/(2p).

But in the second example, the co-efficient also has 2n as its power but when I use 1/(2n) for the root test, it gives me a wrong answer.

So it doesn't look quite the same, isn't it?

I think there's a confusion here: $\displaystyle{\lim\limits_{2p\to\infty}\left(\frac {4^{2p}}{9^pp^{5/4}}\right)^{1/2p}=\frac{4}{\sqrt{9}}\lim\limits_{2p\to\infty}\fr ac{1}{\left(p^{5/4}\right)^{1/2p}}=\frac{4}{3}}$

and all it's fine.

About the second series we get what you said is incorrect, but the series itself is a little weird: why

to write $\displaystyle{\frac{4^{2n}}{2^{2n}}\mbox{ instead of simply }=2^{2n}$ ? Perhaps there's a mistake there?

I think the radius of the series, as you wrote it, must indeed be $\displaystyle{\frac{1}{2\sqrt{e}}}$ ...

Tonio

5. Thanks! Just to clarify a little more. The series converges to $\displaystyle{\frac{1}{2\sqrt{e}}}$ because I took $\frac{1}{2n}$ as the root in the test, ie: $\underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\sqrt{e}}}$.

But if I apply the more generic root in the test, $\frac{1}{n}$, then I will get $\frac{1}{4e}$ as the result, ie: $\underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/n}}=\frac{1}{4e}}$.

Then how I know what I need to take as the root in the test? Can I safely say that as long as I take the power of the $(x+a)$, ie: $(x+\frac{5}{4})^{2n}$, in the test then I will get the correct radius of convergence for the series?

6. Originally Posted by xEnOn
Thanks! Just to clarify a little more. The series converges to $\displaystyle{\frac{1}{2\sqrt{e}}}$ because I took $\frac{1}{2n}$ as the root in the test, ie: $\underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/2n}}=\frac{1}{2\sqrt{e}}}$.

But if I apply the more generic root in the test, $\frac{1}{n}$, then I will get $\frac{1}{4e}$ as the result, ie: $\underset{n\rightarrow\infty}{\lim}\frac{1}{\left[\frac{n^{n^{2}}\cdot e^{3n}}{(2+n)^{n^{2}}2^{2n}}\cdot4^{2n}\right]^{1/n}}=\frac{1}{4e}}$.

Then how I know what I need to take as the root in the test? Can I safely say that as long as I take the power of the $(x+a)$, ie: $(x+\frac{5}{4})^{2n}$, in the test then I will get the correct radius of convergence for the series?

First, it's not the series' but the limit of the inverse of its general term that equals $2e^{-1/2}$ , second

I think this is the actual limit since the root test must be taken with respect to the index of the variable, which

in this case it's x + 5/4 ...

The above is what I think but perhaps someone else could throw some more light into this.

Tonio