Is the series a = [2^n + 3^n]^(1/n) convergent or divergent?

Question

There is supposed to be some type of simplification which makes it easy to see through elementary series, but I can't figure out what it is

Answer 1

You can factor out 3^n; then [2^n+3^n]^(1/n) becomes
(3^n[(2/3)^n + 1^n])^(1/n) = 3[1+(2/3)^n]^(1/n).

Can you take it from there?

Answer 2

Do you mean the series or the sequence?

If it's the sequence, then all you have to do is solve for the limit as n approaches infinity.

lim [2^n + 3^n]^(1/n)
n -> infinity

This is in the form [infinity ^ 0], which means it's indeterminate. We transform the form.

[2^n + 3^n]^(1/n) = e^ ln [ [2^n + 3^n]^(1/n) ]

Use the log property, to get

e^ (ln[2^n + 3^n] / n)

Now we solve for the limit of the power.

lim (ln[2^n + 3^n] / n)
n -> infinity

Using L'Hospital's rule,

lim ( [1/(2^n + 3^n)] [ (2^n)ln2 + (3^n)ln3 ]
n -> infinity

lim { [ (2^n)ln2 + (3^n)ln3 ] / (2^n + 3^n) }
n -> infinity

Divide everything by 3^n.

lim { [ (2/3)^n ln(2) + ln3 ] / [ (2/3)^n + 1 ] }
n -> infinity

Note that as n gets very large, (2/3)^n gets very small, close to 0. Therefore, we can solve the limit directly.

(0 + ln(3)) / (0 + 1)
ln(3) / 1
ln(3)

Note that we originally had e to the power of this, so

e^(ln3)

e and ln are inverses of each other, so

e^(ln3) = 3

The series converges to 3.

Answer 3

ln a = ln(2^n+3^n)/n.
Now this has the behaviout at infinity as
f(x) = ln(2^x+3^x)/x. For this apply L'Hospital and then
lim f(x) = lim (2^x ln 2 + 3^x ln 3 )/(2^x + 3^x) = ln 3.

So ln a = ln 3, a = 3.

I suppose there are other methods too