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Calculus (1)

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Page 1: Calculus (1)

qwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmrtyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiop

MULTI VERIABLE CALCULUS

Assignment

Group Mambers Roll No

Page 2: Calculus (1)

Ratio Test :

Let ∑1

(an ) be a series of positive terms hence suppose that

limn→∞

an+1an

where L is a non negative real number or infinity.

1. If L < 1 , the series Converge.2. If L > 1 or ∞ , the series Diverge.

3. If L = 1 , the test is fail.Test :limn→∞

an+1

an

Question No 1:∑1

( (n+2 )!4 ! n!2n )

an=(n+2 )!4 !n !2n

an+1=(n+1+2 ) !

4 !(n+1)!2n+1

an+1=(n+3 )!

4 !(n+1)!2n+1

limn→∞

an+1

an=

limn→∞

(n+3 )!

4 !(n+1)!2n/(n+2 )!

4 ! n!2n

¿limn→∞

(n+3 ) !

4 ! (n+1 )!2n .2× 4 !n !2

n

(n+2 )!

Page 3: Calculus (1)

¿limn→∞

(n+3 ) (n+2 ) (n+1 )!

4 ! (n+1 )n !2n .2× 4 ! n!2n

(n+2 ) (n+1 ) !

¿limn→∞

(n+3 )

(n+1 )2

¿ 12

limn→∞

n (1+ 3n)

n(1+ 1n)

¿ 12.(1+ 3

∞)

(1+ 1∞

)

¿ 12. 1+01+0

12<1

Converge

ROOT TEST:

Let ∑1

(an ) be a series of positive terms hence suppose that

limn→∞

(an )1n=L where L is a non negative real number or ∞ .

If L < 1 then the series Converge.If L > or ∞ then the series Diverge.

If L = 1 then the test fail.Test :limn→∞

(an)

Page 4: Calculus (1)

Question No 2 :∑1

( n10

)n

an=¿

limn→∞

(an )1n=¿

¿ limn→∞

[ n10

]

¿ [ ∞10

]

¿∞

Diverge