Why you love series, Akul?
Akul is immensely amused by the product of some numbers which follow a particular pattern. Mayank loves to add such "pattern-following-numbers"... Akul devises the following quantity:-
k n = ( n ) . ( n − 2 ) . ( n − 4 ) . . . . . 2 o r 1 ( n − 1 ) . ( n − 3 ) . ( n − 5 ) . . . . . 2 o r 1
Mayank finds the value of lim n ∞ ( e ∑ r = 1 n k 2 r 1 )
Can you guess Mayank's result?
Wanna have more fun with Mayank and Akul. This question is a part of the set Mayank and Akul
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1 solution
Another method:
note that:
π 2 k 2 n = ∫ 0 π / 2 sin 2 n x d x
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Ohh nice Walli's formula!! . Nice :) +1!
I did it like this
That sum is actually equal to Σ k 2 r = ( 1 + x ) m − 1 with x = -1 & m = -1/2 ( when n tends to infinity ) (I derived and then verified also )
Putting values you get sum 0 1 − 1 = ∞ − 1 = ∞
Hence e ∞ 1 = 0
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i dont remember that binomial expansion. but i should rather its important and can be used in many problems
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Yeah! Right! Whenever u see that in a series a new no. Is visible in every next term then 1st check for binomial transformation 👍
you, my friend , r unfortunately wrong ! undefined and infinity r 2 different concepts ! correct me if i'm wrong ! :D
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Its limit , here we deal with tends to not equal to so here 1/0 do tends to infinity... Undefined is the case when we want to equal to as infinity is not defined ! So here everything is fine and correct...
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@Aniket Sanghi – aaah...ask urself is 1/0 is same as very large number such as 10^1000000^1000000000^10000000........... ?
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@A Former Brilliant Member – What are talking man! Check limits you will indeed learn 1/0 tends to infinity similarly 1/infinity tends to 0. @Prakhar Bindal Agreed Bro?
Its a nice one.
Firstly the sum can be transformed into a telescopic series by writing 1 in numerator as (2r+1)-2r.
All the terms get cancelled except the first which is 1 and last crazy term.
For evaluating last term Take log on both sides and try using Riemann Sums . After Using riemann sums you will realise that the sum will diverge towards infinity as it will be some term containing n will be left in numerator which obviously goes towards infinity