Help calculate the limits Help calculate the limits: 
1) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt {k(n-k)}}$$
2) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{(n-k)\ln{n}} $$
3) $$\lim_{n\to \infty}{n}^p \sin(\pi(\sqrt 2 +1)^n) $$
In the directions to this number it is written: compare $$(\sqrt 2 +1)^n$$ with the whole part   $$(\sqrt 2 +1)^n + (-\sqrt 2 +1)^n $$ 
But I just can not understand how this will help. I got that with even n  $$(\sqrt 2 +1)^n + (-\sqrt 2 +1)^n > (\sqrt 2 +1)^n $$
 Аnd with odd on the contrary.
 A: This suppose to be a comment but it getting kinda long so I placed it here in term of an answer instead, although it's not directly answering your question but it may helps... In regard to Rick's comment, I think this is what he meant: 
Suppose I have something like:
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k^2 + n^2}}   $$
Then I can rewrite this as follow:
$$ \lim_{n \rightarrow \infty} \dfrac{1}{ \sqrt{1^2 + n^2}}  + \dfrac{1}{ \sqrt{2^2 + n^2}} + \dfrac{1}{ \sqrt{3^2 + n^2}} + \cdots \dfrac{1}{ \sqrt{n^2 + n^2}}   $$
but note that
$$  \lim_{n \rightarrow \infty} \dfrac{1}{ \sqrt{1^2 + n^2}}  + \dfrac{1}{ \sqrt{2^2 + n^2}}  + \cdots + \dfrac{1}{ \sqrt{n^2 + n^2}} 
 =  \lim_{n \rightarrow \infty} \dfrac{1}{n} \bigg( \dfrac{1}{\sqrt{ \big(\dfrac{1}{n}\big)^2 + 1 }} +  \dfrac{1}{\sqrt{ \big(\dfrac{2}{n}\big)^2 + 1 }}  + \cdots +  \dfrac{1}{\sqrt{ \big(\dfrac{n}{n}\big)^2 + 1 }} \bigg) $$
From here you can see that it starting to look like how you would define a Riemann sum for the function $ \dfrac{1}{\sqrt{x^2 + 1}} $  from $0$ to $1$ with subdivision point $\dfrac{1}{n}, \dfrac{2}{n}, \cdots $  So now you can evaluate 
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k^2 + n^2}} = \int_0^1 \dfrac{1}{\sqrt{x^2 + 1}} dx = \log \big[ x + \sqrt(x^2 + 1) \big] \bigg|_0^1 = \log(1 + \sqrt{2} )   $$ 
For your problem, can you think of doing something similar? Hope this help... 
