Power series: radius of convergence

[DottZakapa]

Homework Statement

radius of convergence of the following power series
$\sum_{k=0}^\infty \frac {2^n+3^n}{4^n+5^n} x^n$

Relevant Equations

convergence tests

$\sum_{k=0}^\infty \frac {2^n+3^n}{4^n+5^n} x^n$

in order to find the radius of convergence i apply the root test, that is

$\lim_{n \rightarrow +\infty} \sqrt [n]\frac {2^n+3^n}{4^n+5^n}$

$\lim_{n \rightarrow +\infty} \left(\frac {2^n+3^n}{4^n+5^n}\right)^\left(\frac 1 n\right)=\lim_{n \rightarrow +\infty} \left(\frac {3^n \left(\frac {2^n}{3^n}+1\right)}{5^n\left(\frac{4^n }{5^n}+1\right)}\right)^\left(\frac 1 n\right)= \lim_{n \rightarrow +\infty} \left(\frac {3} {5}\right) \left(\frac {\left(\left(\frac {2} {3}\right)^n +1\right)} {\left(\left(\frac{4 } {5}\right)^n+1 \right)}\right)^\left (\frac 1 n \right)=\left(\frac {3} {5}\right) \lim_{n \rightarrow +\infty} \left(\frac {\left(\left(\frac {2} {3}\right)^n +1\right)^\left (\frac 1 n \right)} {\left(\left(\frac{4 } {5}\right)^n+1 \right)^\left (\frac 1 n \right)}\right)$

at this point i am stuck, don't know how to handle it, probably i am forgetting some properties in order to simplify, supposing that up to here i did it correctly.

 

[scottdave]

How does the limit of the denominator behave? How about the limit of the numerator?

 

[DottZakapa]

$\lim_{n \rightarrow +\infty}e^\left(\frac {log\left(1+\left(\frac 2 3\right)^n\right)}{n}\right)=\lim_{n \rightarrow +\infty}e^\left(\left(\frac 2 3\right)^n\frac 1n\right)$

in which

$\lim_{n \rightarrow +\infty}\left(\frac 2 3\right)^n=0$

$\lim_{n \rightarrow +\infty}\frac 1n=0$

hence, putting it all together

$\lim_{n \rightarrow +\infty}e^\left(0\right)= e^0=1$

then same procedure for the denominator

is all this correct ?

 

[FactChecker]

I think that @scottdave is hinting at something simpler. Which term in the numerator dominates for large n? Ignore the other term. Same for the denominator.

 

[Mark44]

Homework Statement:: radius of convergence of the following power series
$\sum_{k=0}^\infty \frac {2^n+3^n}{4^n+5^n} x^n$
Relevant Equations:: convergence tests

$\sum_{k=0}^\infty \frac {2^n+3^n}{4^n+5^n} x^n$

in order to find the radius of convergence i apply the root test, that is

Minor point, but the summation should be on n rather than k, or else all the exponents should be k.

$\lim_{n \rightarrow +\infty} \sqrt [n]\frac {2^n+3^n}{4^n+5^n}$

The limit should be applied to the whole n-th term. IOW, including $x^n$.

$\lim_{n \rightarrow +\infty} \left(\frac {2^n+3^n}{4^n+5^n}\right)^\left(\frac 1 n\right)=\lim_{n \rightarrow +\infty} \left(\frac {3^n \left(\frac {2^n}{3^n}+1\right)}{5^n\left(\frac{4^n }{5^n}+1\right)}\right)^\left(\frac 1 n\right)= \lim_{n \rightarrow +\infty} \left(\frac {3} {5}\right) \left(\frac {\left(\left(\frac {2} {3}\right)^n +1\right)} {\left(\left(\frac{4 } {5}\right)^n+1 \right)}\right)^\left (\frac 1 n \right)=\left(\frac {3} {5}\right) \lim_{n \rightarrow +\infty} \left(\frac {\left(\left(\frac {2} {3}\right)^n +1\right)^\left (\frac 1 n \right)} {\left(\left(\frac{4 } {5}\right)^n+1 \right)^\left (\frac 1 n \right)}\right)$

at this point i am stuck, don't know how to handle it, probably i am forgetting some properties in order to simplify, supposing that up to here i did it correctly.

As already noted, the fraction behaves the same as $\frac {3^n}{5^n}$ in the limit.

Also, you seem to automatically pick the Root Test, at least in two or three threads I've seen. Other tests, such as the Ratio Test, can sometimes be simpler to apply.

 

[DottZakapa]

in the numerator dominates $3^n$ in the denominator $5^n$$\Rightarrow \lim_{n \rightarrow +\infty} \sqrt [n]\frac {2^n+3^n}{4^n+5^n}\sim \lim_{n \rightarrow +\infty}$$\sqrt [n]{ \left(\frac 3 5 \right)^n} \sim \frac 3 5$

correct?

 

[Mark44]

in the numerator dominates $3^n$ in the denominator $5^n$$\Rightarrow \lim_{n \rightarrow +\infty} \sqrt [n]\frac {2^n+3^n}{4^n+5^n}\sim \lim_{n \rightarrow +\infty}$$\sqrt [n]{ \left(\frac 3 5 \right)^n} \sim \frac 3 5$

correct?

Yes, but again, you need to include $x^n$ in the limit. Otherwise you won't be able to get the interval of convergence.

 

[DottZakapa]

Yes, but again, you need to include xnxnx^n in the limit. Otherwise you won't be able to get the interval of convergence.

given

$\sum_{n=0}^\infty \frac {2^n+3^n}{4^n+5^n} x^n$

the professor showed us to see it as

$\sum_{n=0}^\infty a_n x^n$

and then study $a_n$ as you have seen i do.

Concerning the x in case x is not centred at 0 then i use the substitution $t= (x-x_0)$ or what ever equation, then, once found the radius.|t|<R

-R<t<R

then study the boundaries

Don't know if i made my self clear

This is why i repeatedly solve in this way

 

[Mark44]

Here's a slightly simpler problem ( so that I don't have to write so much), but with exactly the same interval of convergence -- $\sum_{n = 0}^\infty \frac{3^n x^n}{5^n}$
Using the Ratio Test, I have
$\lim_{n \to \infty}\frac{3^{n + 1} |x|^{n + 1}}{5^{x + 1}} \cdot \frac{5^n}{3^n |x|^n}$
$= \frac{3|x|} 5$

For absolute convergence, we must have $\frac{3|x|} 5 \Rightarrow |x| < \frac 5 3$
or $-\frac 5 3 < x < \frac 5 3$
The only things remaining are to check the two endpoints of the interval for convergence.