The value of 1.5-0-8 is.
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We have for x∈R,
x−1<[x]≤x
or, nx−1<[nx]≤nx [For n∈N].
Then,
(x−1)+(2x−1)+....+(nx−1)<[x]+[2x]+....+[nx]≤x+2x+.....+nx
⇒
2
n(n+1)
x−n<[x]+[2x]+....+[nx]≤
2
n(n+1)
x
⇒(
2
1
+
2n
1
)x−
n
1
<
n
2
[x]+[2x]+....+[nx]
≤(
2n
1
+
2n
1
)x
⇒
n→∞
lim
(
2
1
+
2n
1
)x−
n
1
≤
n→∞
lim
n
2
[x]+[2x]+....+[nx]
≤
n→∞
lim
(
2
1
+
2n
1
)x [Using limit property]
⇒
2
x
≤
n→∞
lim
n
2
[x]+[2x]+....+[nx]
≤
2
x
Using Sandwich property we get,
n→∞
lim
n
2
[x]+[2x]+....+[nx]
=
2
x
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