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kishan12

_{[ n]}

_{[ inf]} {1/n^2 sec^2(1/n^2) + 2/n^2 sec^2(4/n^2) + .......1/nsec^2(1)}

Rates assured

Ans is 1/2tan(1)

karthik2007

Sum is : (I am not writing the limits.. too lazy to type :P)

(1/n² sec²(1/n²) + 2/n² sec²(4/n²) + ....) = 1/n(1/n sec²(1/n²) + 2/n sec²(4/n²) + 3/n Sec²(9/n²) +.....)

Thus, the general term is :

(r=1 to n) 1/n(r/n sec²(r/n)²)

= _[0]

^[1] xsec²(x²) dx. Take x²=t => dt = 2x.

= _[0]

^[1] sec²tdt/2 = tant/2]¹_{0 =}(tan1 - tan0)/2 = (tan1)/2. Hence the answer

karthik2007

Here are some useful formulae regarding limits that I know:

1.lt x-> infinity

(r=1 to n) (1/n)f(r/n). If you get the general term of a given sum (Just like the one that you gave), here is how you proceed:

Put 1/n as dx. Put r/n as x. Take the limits as 0, 1. Hence, the answer in this case is: _[0]

^[1] f(x)dx.

2. Limit n-> infinity(1^p+2^p+3^p+......n^p/n^p+1) = 1/p+1.

3. limit x-> infinity [ax+b]/x = a. []-> step function.

4. _{[x ]}

_{[infinity ]} a₀x^p + a₁x^p-1 + a₂x^p-2..../b₀x^q+b₁x^q-1... = 0, if p<q.
=a₀/b₀, if p=q, and is equal to infinity if p>q.

5._{[x ]}

_{[infinity ]} (x²+ax+b/x²+bx+c)^x+f = e^a-c..

Now, try solving these problems:

1._{[n ]}

_{[infinity ]}

(r=0 to n) 1/n sin(r

/2n)
2. _{[x ]}

_{[infinity ]} [6x+7]/x
3._{[n ]}

_{[infinity ]}

(r=0 to n-1) e^r/n/n.

Hope you understood the concept.

kishan12

just give me 5 min
i will solve the first one
2and one its 6

karthik2007

Good. You are right. Give me the answer for the remaining two. Remember, 1/n = dx and r/n = x, and then normal integration. Remove the sigma signs after you make these substitutions, and also remove the limits. Just do integration between these limits.

kishan12

limit x-> infinity [ax+b]/x = a. []-> step function.
what is Step fuction here ??

karthik2007

the function enclosed within the square brackets, ie [ ] is a step function.

If you did not know what a step function is, here goes:

A step function basically returns the integer lower than or equal to the value of x that is there in the square bracket.

for example, for a function f(x) = [x], if x = 4.98, [x] = [4.98] = 4 (Integer lower or equal to).

[5.66] = 5.
[4] = 4.

kishan12

_{[ 0]}

^{[ 1]} sin(pi x/2)dx = cos(pi x /2)*x^2/2 =0

Hope did the integration part correctly

karthik2007

Umm.... you are giving the answer that I gave.... I hope you know integration? are you in 11th? anyway, let me give you these:

integral sin(mx) = -cos(mx)/m. integral cos(mx) = -sin(mx)/m.

here, it is integral sin(x pi/2) = (here m = pi/2) -cos(xpi/2)/pi/2.

After you integrate, you first put the value of x as 0, then put the value of x as 1 (as they are the lower and upper limits), and then subtract the two quantities.

I suggest you forget these type of problems for now if integration has not been done in school.

kishan12

third one is it (e - 1)

karthik2007

BINGO! Right on the money! Way to go pal. I would be happy if you have understood the steps. Have fun!

karthik2007

Here, take my rates for solving it correctly...

karthik2007

I want you to try the problem that you posted here yourself.

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