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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 Jul 2007 17:19:40 IST
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find the limit for n approaching infinity...
1/(n3+1) + 4/(n3+1) +.............+n2/(n3+1)
if we take 1/(n3+1) common and apply the formula for summation of k2,the answer comes out to be 1/3...dats given in the solution for all such type of questions...
But when n approaches infinity...all terms become(approach) 0...for the last term...bring n2 in Dr and it becomes 1/(n+1/n2)...so when n approaches infinity....it also approaches 0....so answer is 0!
plzz. clear my confusion...
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15 Jul 2007 17:59:11 IST
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[n ] [ infinity] [r = 1] [r = n ] r2 / (n3 +1)
= [n ] [ infinity] [ n (n+1) (2n + 1)] / [ 6. (n3 + 1) ]
= [n ][ infinity] [ n (n+1) (2n + 1) ] / [ 6 . (n+1) . (n2 - n + 1) ]
= [n ][ infinity] [ n (2n + 1) ] / [ 6 . (n2 - n + 1) ]
= [n ][ infinity] [ 2 + 1 / n ] / [ 6 . ( 1 - 1 / n + 1 / n2 ) ]
= 2 / 6 = 1 / 3
Now, as said by Neeraj, we see that each term individually [n ][ infinity] r / (n3 + 1) tends to zero. But when such small positive quantities in infinite numbers are added the result tends to 1/3 as obtained. Remember that the infinite sum of infinitesimally small positive quantities will tend to a finite value.
Hope that removes all your doubt.
Cheers !
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You never know what is enough till you know what is more than enough.
Titun |
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15 Jul 2007 18:02:29 IST
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oh ya..
thanx a lot titun!
i never thought of it...
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15 Jul 2007 18:08:09 IST
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is it used in every question in which involves no. of terms approaching infinity?
so i shud be careful in all such question...
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