Byju's Answer
Standard XII
Mathematics
Definite Integral as Limit of Sum
limn →∞ [ 1/n...
Question
l
i
m
n
→
∞
[
1
n
+
1
+
1
n
+
2
+
⋯
1
n
+
n
]
is equal to
A
3 log 2
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B
log 2
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C
2 log 2
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D
None of these
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Solution
The correct option is
B
log 2
l
i
m
n
→
∞
[
1
n
+
1
+
1
n
+
2
+
⋯
1
n
+
n
]
=
l
i
m
n
→
∞
∑
n
r
=
1
1
n
+
r
=
l
i
m
n
→
∞
∑
n
r
=
1
1
n
.
1
1
+
r
n
=
∫
1
0
1
1
+
x
d
x
=
[
l
o
g
(
1
+
x
)
]
1
0
=
l
o
g
2.
Suggest Corrections
0
Similar questions
Q.
The sum of the series
1
log
2
4
+
1
log
4
4
+
1
log
8
4
+
.
.
.
.
+
1
log
2
n
4
is
(a)
n
(
n
+
1
)
2
(b)
n
(
n
+
1
)
(
2
n
+
1
)
12
(c)
n
(
n
+
1
)
4
(d) none of these
Q.
l
i
m
n
→
∞
[
1
n
+
1
+
1
n
+
2
+
⋯
1
n
+
n
]
is equal to
Q.
l
i
m
n
→
∞
{
l
o
g
n
−
1
(
n
)
l
o
g
n
(
n
+
1
)
l
o
g
n
+
1
(
n
+
2
)
…
…
l
o
g
(
n
k
−
1
)
(
n
k
)
}
is equal to
Q.
If
N
=
n
!
(
n
∈
N
,
n
>
2
)
, then
lim
N
→
∞
[
(
log
2
N
)
−
1
+
(
log
3
N
)
−
1
+
.
.
.
.
.
(
log
n
N
)
−
1
]
is
Q.
What is
1
log
2
N
+
1
log
3
N
+
1
log
4
N
+
.
.
.
.
.
.
1
log
100
N
equal to
(
N
≠
1
)
?
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