This, as you would see if you knew romanian, is an exercise given

>>8471684
This shit for* even though i think that shitlove sounds way better

ln(3)

>>8471684
ln(3) is an upper bound

>>8471684

Factor out [math] 3^n [/math], [eqn] \ln \left ( 3^n \left [ 1+3^{1/2} + 3^{1/3} + \cdots + 3^{1/n} \right ] \right ) \\ = \frac { \ln (3^n) } { n } + \frac { 1 } { n } \ln \left [ 1+3^{1/2} + 3^{1/3} + \cdots + 3^{1/n} \right ] \\ = \ln ( 3 ) + \frac { 1 } { n } \ln \left [ 1+3^{1/2} + 3^{1/3} + \cdots + 3^{1/n} \right ] [/eqn] Now from here use the fact that [math] 1 + 3^{1/2} + \cdots + 3^{1/n} [/math] is monotonic and bounded (that fact it's monotonic is trivial, however you'll need to prove that it's bounded, I just assumed that it was), then by the monotone convergence theorem it converges to it's least upper bound. Clearly then the right hand side tends to 0 and you're left with just [math] \ln ( 3 ) [/math]

At least that's how I think it's done. You should verify the individual steps for yourself, in particular my assumption of boundedness.

>>8471684
1/nlog(3^n+...+3)-log(3)=1/n(log(3^n+...+3)-nlog(3))
=1/n log({3^n+...+3}/3^n)
=1/n log(1+...+3^{1-n})
=1/n log({1-1/3^n}/{1-1/3})
=1/n{log(1-1/3^n)-log(2/3)} -> 0

>>8471768
>[math]3^n 3^{1/2} = 3^{n/2}[/math]
Full retard

>>8471684
[math] \frac{1}{n} \ln (3^{n}) < a_n < \frac{1}{n} \ln (3^{n} + 3^{n-1} + 3^{n-2} + \cdots + 3^{0}) < \frac{1}{n} \ln (3^{n+1}) [/math]

>>8471809
thanks bro. I didn't realise that the first term was so bloody obvious

>>8471809
And how do you prove that [math]\sum_{k \,=\, 1}^n 3^\frac{n}{k} \leqslant \sum_{k \,=\, 1}^n 3^{n \,-\, k}[/math]?

>>8471865
yo don't even have to do so. put n*3^n and it still werks

>>8471865
[math] 3^{\frac{n}{k+1}} < 3^{n-k} [/math] for all [math] n > k + 1 [/math]

>>8471871
because the nth root of n is 1.

>>8472267
if you take the limit it is :)

>>8472325
indeed, I made a semi-brainfart there.