Question

Find the sum of the infinite series $\frac{2}{\lfloor 1}+\frac{12}{\lfloor 2}+\frac{28}{\lfloor 3}+\frac{50}{\lfloor 4}+\frac{78}{\lfloor 5}+.....$

Answer 1

The $n^{th}$ term of the series $2,12,28,50,78,....$ is $3n^2+n-2;$ hence
$u_n=\frac{3n^2+n-2}{\lfloor n}=\frac{3n(n-1)+4n-2}{\lfloor n}$
$=\frac{3}{\lfloor n-2}+\frac{4}{\lfloor n-1}\frac{2}{\lfloor n}$.
Put $n$ equal to $1,2,3,4,...$ in succession$;$ then we have
$u_1=4-\frac{2}{\lfloor 1};$ $u_2=3+\frac{4}{\lfloor 1}-\frac{2}{\lfloor 2};$ $u_3=\frac{3}{\lfloor 1}+\frac{4}{\lfloor 2}-\frac{2}{\lfloor 3};$
and so on.
Hence, $S_{\infty }=3e+4e-2(e-1)=5e+2$.