Mathematical Reflections 2017, Issue 1 - Problem S402

Problem:
Prove that $$\sum_{k=1}^{31} \dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}<\dfrac{3}{2}+\sum_{k=1}^{31} (k-1)^{1/5}.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}<\dfrac{k}{3(k-1)^{4/5}},$$ so
$$\begin{array}{lll} \displaystyle \sum_{k=1}^{31} \left(\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}-(k-1)^{1/5}\right)&<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\sum_{k=2}^{31} \dfrac{2k-3}{(k-1)^{4/5}} \\ &<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\sum_{k=2}^{31} \dfrac{2k-3}{k-1}\\&<& \displaystyle \dfrac{1}{1+2^{4/5}}-\dfrac{1}{3}\left(1+\dfrac{3}{2}\right)<0. \end{array}$$
So, $$\sum_{k=1}^{31} \left(\dfrac{k}{(k-1)^{4/5}+k^{4/5}+(k+1)^{4/5}}-(k-1)^{1/5}\right)<0<\dfrac{3}{2}.$$