Limiting sum help! (1 Viewer)

[plshelp]

I just need help in finding the limiting sume:

1/4 + 1/4*7 + 1/7*10 +...+ 1/(3n-2)(3n+1) = n/(3n+1)

thanx heaps

 

[random-1006]

I just need help in finding the limiting sume:

1/4 + 1/4*7 + 1/7*10 +...+ 1/(3n-2)(3n+1) = n/(3n+1)

thanx heaps

This is mathematical induction, NOT LIMITING SUM

ill assume we are proving true for n>= 1

now prove true n=1 { sub n=1 into general term ( last term) of LHS and RHS } , they are equal

Assume the statement true for n=k

ie S (k) = k / ( 3k+1)

Now we try to prove it true n=k+1

ie we try to prove that S (k+1) = (k+1) / ( 3(k+1) + 1 ) = ( k+1) / ( 3k +4)

Now S (k+1) = S (k) + T (k+1) LEARN THIS RESULT WELL

ie the sum of 8 terms would equal the sum of the first 7 , plus term number 8

so S (k+1) = S ( k) + T (k+1)

now from assumption we have S (k) = k / ( 3k+1)

and to find T ( k+1) we sub "k+1" into the general term on the LHS

so S (k+1) = k / (3k+1) + 1/ ( 3(k+1) -2 ) ( 3(k+1) +1)
S (k+1) = k/ ( 3k+1) +1/( 3k+1)( 3k+4)=[ k(3k+4) +1 ] / ( 3k+1) ( 3k+4)

= [ 3k^2 +4k +1] / [ (3k+1)(3k+4) ]

now factorise top , two numbers with product of 3 and sum of 4

= [ 3k^2 +3k +k +1 ] / [ (3k+1) ( 3k+4) ]
= [3k(k+1) + 1 ( k+1) ] / [ (3k+1) ( 3k+1) ]
=[ ( 3k+1) ( k+1)] / [ (3k+1) ( 3k+4) ]
= (k+1) / ( 3k+4)

Conclusion. If it is true for n=k, then it is true for n=k+1, since it true for n=1, it is true for n=2, 3,4 and so on

Therefore it is true for all integers n >=1

NOTE : S (k+1) means sum of k+1 terms , it doesnt not mean S times (k+1), similarly with T (k+1) , the parts in the brackets are subscript .

General steps

Step 1 : prove true for the starting case ( usually n=1, but not always)

Step 2 Assume its true for n=k

Step 3: Use your assumption from step 2 to prove true n=k+1 ( for series questions use the result S (k+1) = S (k) + T (k+1)

Step 4 : Conclusion, it doesnt have to be the same word for word that i have put, but something similar. It is merely a formality.

 

[plshelp]

No! i know its a mathematical induction! But there was a question after that that says now that you've proved it, find the limiting sum...

thats what i dont understand

 

[random-1006]

No! i know its a mathematical induction! But there was a question after that that says now that you've proved it, find the limiting sum...

thats what i dont understand

ok so we have the RHS, now its saying take the limit as n--> infinity of RHS

so limit n--> infinity of [ n/ ( 3n+1)] , divide by highest power n in denominator

so 1/ ( 3 + 1/n) which goes to 1/3
ANSWER=1/3

 

[hscishard]

1.

 

[nat_doc]

 

[hscishard]

sht no.

Divide n/n
=1/[3 + 1/0]

n--> infinity

=1/3

 

[plshelp]

oh...thanx so much:wave: now i get it...XD

 

[random-1006]

sht no.

Divide n/n
=1/[3 + 1/0]

n--> infinity

=1/3

whats this lol, you cnt divide by zero, you mean 1/ infinity , which goes to zero

 

[hscishard]

Crap should be 1/n

 

[Drongoski]

I just need help in finding the limiting sume:

1/4 + 1/4*7 + 1/7*10 +...+ 1/(3n-2)(3n+1) = n/(3n+1)

thanx heaps

Unfortunately halfway thru my solution LaTeX conked out on me; so now without it.

If you want to derive the formula for the sum of the series, here is 1 way:



1/(3n-2)(3n+1) = 1/3[1/(3n-2) - 1/(3n+1)]

Let f(n) = 1/3(3n-2)

.: 1/(3n-2)(3n+1) = f(n) - f(n+1)

.: the series = [f(1)-f(2)] + [f(2)-f(3)] + [f(3)-f(4)] + . . . . [f(n)-f(n+1)]

= f(1) - f(n+1) = 1/3(3x1 -2) -1/3(3(n+1)-2)

= 1/3[1 - 1/(3n+1)] = 3n/3(3n+1)

= n/(3n+1)



which goes to 1/3 as n --> infinity as pointed out above.

 

[Drongoski]

In LaTeX