lim x->0 x([1/x]+[2/x]...+[k/x])by using sandwich theorem, i am getting k/2 on left and k^2/2 on right. can someone please help me with this.thank you so much *

Samyak Jain
333 Points
3 years ago
According to me, both mentioned values are wrong.
Greatest integer is always less than or equal to the number.
x – 1  $\dpi{80} <$  [x]  $\dpi{80} \leq$  x
Now, limx$\dpi{80} \rightarrow$0 x([1/x] + [2/x] + … + [k/x]) can be written as limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ ([1x] + [2x] + … + [kx]) / x
(x –1) + (2x –1) + … + (kx –1) $\dpi{80} <$ ([1x] + [2x] + … + [kx]) $\dpi{80} \leq$ x + 2x + … + kx
$\dpi{80} \Rightarrow$ {k(k + 1)/2}/x  –  k  $\dpi{80} <$  ([1x] + [2x] + … + [kx])  $\dpi{80} \leq$  {k(k + 1)/2}/x
Divide throughout by x and take limit as x$\dpi{80} \rightarrow$$\dpi{80} \infty$.
$\dpi{80} \Rightarrow$ limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ {k(k + 1)/2}/x  –  k  $\dpi{80} <$  limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ ([1x] + [2x] + … + [kx])  $\dpi{80} \leq$  limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ {k(k + 1)/2}/x
As limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ {k(k + 1)/2}/x  –  k  =  limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ {k(k + 1)/2}/x  =  k(k+1)/2,
by sandwich theorem, limx$\dpi{80} \rightarrow$$\dpi{80} \infty$ ([1x] + [2x] + … + [kx]) = k(k+1)/2
i.e.  limx$\dpi{80} \rightarrow$0 x([1/x] + [2/x] + … + [k/x])  =  k(k+1)/2