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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper
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© examsnet.com
Question : 76
Total: 160
lim
n
→
∞
√
1
+
√
2
+
.
.
.
+
√
n
n
3
2
=
0
2
3
1
3
2
Validate
Solution:
The expression is given as,
I
=
lim
n
→
∞
√
1
+
√
2
+
.
.
.
.
+
√
n
n
3
2
=
lim
n
→
∞
√
1
+
√
2
+
.
.
.
.
+
√
n
n
√
n
=
lim
n
→
∞
1
n
[
√
1
n
+
√
2
n
+
.
.
.
.
+
√
n
n
]
=
n
∑
r
=
1
lim
n
→
∞
1
n
√
1
n
=
1
∫
0
√
x
d
x
=
2
3
[
x
3
2
]
0
1
=
2
3
© examsnet.com
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