Can Calculus Tricks Simplify These Limit Problems?

[amcavoy]

Can anyone help with these?

1. \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{k}}{n}

2. \lim_{n\rightarrow 0}\frac{\log_{2}\sum_{k=1}^{2^n}\sqrt{2k-1}}{n}

Thanks for you help.

 

[saltydog]

Regarding just the first one:

Although I question this, Mathematica returns:

\mathop \lim\limits_{n\to 0}\frac{Ln[\sum_{k=1}^{2^n}\sqrt{k}]}{n}\approx 0.8527

Seems to me that it should be -\infty

Since once n gets below 1, the sum goes to just 1.

How about this also too?

\mathop \lim\limits_{n\to \infty}\frac{Ln[\sum_{k=1}^{2^n}\sqrt{k}]}{n}

This one Mathematica returns 1.0397 which makes sense if you plot the values for a range of n, it seems to approach a value near this.

I'd like someone to explain these also.

 

[amcavoy]

I encountered these problems on another site, and was just interested in them. I know some single-variable calculus, but I am just working on sequences and series now, so I was wondering if anyone knew of a trick to get these done.

Thanks for your help.