[math]\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{k}}{n}[/math]

 

[math]\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{2k-1}}{n}[/math]

 

I have had people give me the results after plugging it into Mathematica/Maple, but I was wondering if anyone knew how to solve these without a computer.

I wouldn´t even see how the terms are defined for non-integer values of n. E.g. n=0.5, what´s the sum from k=1 to k=sqrt(2)? Even when I rewrite the logarithm of a sum to a product of logarithms (that´s what 1st comes to my mind when I see such terms) a non-integer number of factors doesn´t seem to make sense at a first glance.

[math]\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{k}}{n}[/math]

 

 

At n=0' date=' you have:

 

[math']\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{1}\sqrt{k}}{0}[/math]

 

[math]\displaystyle \lim_{n\rightarrow 0}\frac{\log_{2}(1)}{0}[/math]

 

[math] log_b(x) = \frac{lnx}{lnb} [/math]

 

[math] log_2(1) = \frac{ln 1}{ln2} [/math]

 

The natural log of 1 is zero, and the natural log of 2 is around .69, so:

 

[math]\displaystyle \lim_{n\rightarrow 0}\frac{0}{0}[/math]

 

What does n range over?

I'm not sure, that is all the information that was provided. Someone else plugged it into Mathematica and got [math]\frac{3}{2}[/math]. They used the inequality [math]\int_{1}^{2^n}\sqrt{x} dx < \sum_{k=1}^{2^n}\sqrt{k} < 1+\int_{1}^{2^n}\sqrt{x} dx[/math].

 

I came across these problems on the internet. I just wanted to know if I'm missing something (being in Calculus) and not knowing these.