Prove by induction - can not prove the step case I have to prove by induction the following:
$$\sum^{n-1}_{k=2} k\log_{10} k \le \frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2$$
I have proven the base case, but I am stuck at the step case. I have tried different ways but always have stuck somewhere and could not finish it.
Anyone has any idea?
Thank you
 A: What you want to show is that
$\sum^{n-1}_{k=2} k\log_{10} k 
\le \frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2
$
implies
$\sum^{n}_{k=2} k\log_{10} k 
\le \frac{1}{2}(n+1)^2\log_{10} (n+1)-\frac{1}{8}(n+1)^2
$.
From your assumption,
$\begin{array}\\
\sum^{n}_{k=2} k\log_{10} k 
&=\sum^{n-1}_{k=2} k\log_{10} k+n\log_{10} n\\
&\le\frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2+n\log_{10} n\\
\end{array}
$ 
Therefore,
you have succeeded
if you can prove
$\frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2+n\log_{10} n
\le \frac{1}{2}(n+1)^2\log_{10} (n+1)-\frac{1}{8}(n+1)^2
$.
I will leave this up to you.
(Since you asked for help.
here it is.
Note that this is just algebra.)
You want to show that
$\begin{array}\\
0
&\le \frac{1}{2}(n+1)^2\log (n+1)-\frac{1}{8}(n+1)^2
-\frac{1}{2}n^2\log n+\frac{1}{8}n^2+n\log n\\
&= \frac{1}{2}(n^2+2n+1)\log (n+1)-\frac{1}{8}(n^2+2n+1)
-\frac{1}{2}n^2\log n-\frac{1}{8}n^2+n\log n\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+\frac{1}{2}(2n+1)\log (n+1)-\frac{1}{8}(2n+1)+n\log n\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+(n+\frac12)\log (n+1)-\frac{1}{8}(2n+1)+n\log n\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+n(\log (n+1)+\log (n))+\frac12\log (n+1)-\frac{1}{8}(2n+1)\\
&= \frac{1}{2}n^2(\log (n+1)-\log (n))+n(\log (n+1)+\log (n)-\frac14)+\frac12\log (n+1)-\frac{1}{8}\\
\end{array}
$
and all these terms
are positive
for 
$n \ge 2$.
