# Thread: math homework help: infinite series

1. ## math homework help: infinite series

Hello!

I've been doing math homework and am going over the problems I had difficulty completing. I'm on infinite series. I'd greatly appreciate a walkthrough/explanation for the following problem and some tips on how to approach problems like this/proof problems in general.

Evaluate: **"<=" means "less than or equal to"; "An" means "Ace of n"

lim ((2^n)+(3^n))^(1/n) Hint: Show that 3<=An<=(2 x 3^n)^(1/n)
n-->infinite

This is an even problem, so I do not have a solution. I tried solving it the easy way, by which I mean setting n equal to infinite, making the equation inside to the zero power, which in turn makes the limit 1. However, because I did not use the "hint," I fear that I may have done something wrong.

Thank you!

2. Originally Posted by wyhwang7
Hello!

I've been doing math homework and am going over the problems I had difficulty completing. I'm on infinite series. I'd greatly appreciate a walkthrough/explanation for the following problem and some tips on how to approach problems like this/proof problems in general.

Evaluate: **"<=" means "less than or equal to"; "An" means "Ace of n"

lim ((2^n)+(3^n))^(1/n) Hint: Show that 3<=An<=(2 x 3^n)^(1/n)
n-->infinite
This is a sequence problem not a series problem!

$\displaystyle 3^n \leq 2^n + 3^n \leq 3^n + 3^n = 2\cdot 3^n \implies 3 \leq \left( 2^n + 3^n\right)^{1/n} \leq 3 \cdot 2^{1/n}$
Now use the squeeze theorem.

This problem has a simple generalization. Let $\displaystyle a_1,...,a_k$ be positive real numbers.
Then $\displaystyle (a_1^n+...+a_k^n)^{1/n} \to \max\{ a_1,...,a_k\}$. Try to see if you can prove it.

3. Originally Posted by ThePerfectHacker
This is a sequence problem not a series problem!

$\displaystyle 3^n \leq 2^n + 3^n \leq 3^n + 3^n = 2\cdot 3^n \implies 3 \leq \left( 2^n + 3^n\right)^{1/n} \leq 3 \cdot 2^{1/n}$
Now use the squeeze theorem.

This problem has a simple generalization. Let $\displaystyle a_1,...,a_k$ be positive real numbers.
Then $\displaystyle (a_1^n+...+a_k^n)^{1/n} \to \max\{ a_1,...,a_k\}$. Try to see if you can prove it.
Thank you for your help! I see how you arrived at
3 <= (2^n + 3^n)^(1/n) <= (2 x 3^n)^(1/n) by raising everything to the power of 1/n.

Using the Squeeze Theorem, would that make the limit 3?

Oh, and how did you know to use 3^n <= 2^n + 3^2 <= 3^n + 3^n

That is, how did you derive the values of Bn and Cn?

4. Originally Posted by wyhwang7
Thank you for your help! I see how you arrived at
3 <= (2^n + 3^n)^(1/n) <= (2 x 3^n)^(1/n) by raising everything to the power of 1/n.

Using the Squeeze Theorem, would that make the limit 3?
Yes . This is because $\displaystyle 2^{1/n}\to 1$.

Oh, and how did you know to use 3^n <= 2^n + 3^2 <= 3^n + 3^n

That is, how did you derive the values of Bn and Cn?
I just thought of what I can use to simplify the expression $\displaystyle 2^n+3^n$. And I saw that if you forget about the $\displaystyle 2^n$ then the expression because $\displaystyle 3^n$ which is easier to work with than $\displaystyle 2^n+3^n$. Likewise when I wrote $\displaystyle 2^n+3^n\leq 2\cdot 3^n$, again it made the expression easier to work with being under $\displaystyle 1/n$ power.