Upper bound for expression involving logarithms Let $N = 2^p$ for some $p \in \mathbb{N}$. Find the smallest upper bound for $\frac{N}{2}\log\left(\frac{N}{2}\right) + \frac{N}{4}\log\left(\frac{N}{4}\right) + \ldots + 1$
I guess I could first rewrite this to $\frac{2^p}{2}\log\left(\frac{2^p}{2}\right) + \frac{2^p}{4}\log\left(\frac{2^p}{4}\right) +\ldots+ 1$ and then to 
$2^{p-1}\log(2^{p-1}) + 2^{p-2}\log(2^{p-2}) +\ldots+1$ but I still don't know how I should proceed.
All help appreciated.
Edit: Now I though also writing it to the form $(p-1)2^{p-1} + (p-2)2^{p-2} + ... + 1 \Leftrightarrow \sum_{i=1}^{log N} (p-i)2^{p-i}$ but I still feel like ......  ;-( (yes, the log is base 2, sorry, forgot to mention that)

Fix $k = \log 2 = 1$. Then expression yields 
$$\begin {eqnarray}
S & = & \frac {N} {2} (\log N - k) + \frac {N} {4} (\log N - 2k) + \cdots + \frac {N} {2^{p - 1}} (\log N - (p - 1)k) + 1 \nonumber \\
& = & N \log N (\frac {1} {2} + \frac {1} {4} + \cdots + \frac {1} {2^{p - 1}}) - N (\frac {1} {2} + \frac {2} {4} + \cdots + \frac {p - 1} {2^{p - 1}}) + 1 \nonumber \\
& > & N \log N - 2 \log N - 2N + 1 \nonumber \\
& = & p 2^p - 2^p - 2p + 1.
\end {eqnarray}$$
