Question

\[f(x)=\sum_{\lambda=1}^{n}(x-\frac{ 1 }{ \lambda })(x-\frac{ 1 }{ \lambda+1 })\]then \[\lim_{n \rightarrow \infty}f(0)=?\]

Answer 1

@ganeshie8 ?

Answer 2

@Miracrown ?

Answer 3

@mathslover ?

Answer 4

Oh another one.. I hope that this time, it will be easy :P

Okay, give me time, I will try.

Answer 5

ok

Answer 6

brb in 5 minutes.

Answer 7

@dan815

Answer 8

You need to solve this :
\(\cfrac{1}{2} + \cfrac{1}{6} + \cfrac{1}{12} + \cfrac{1}{20} + ... \)

Answer 9

I got the answer... let wolfram verify it first :)

Answer 10

After solving that, I get 1 as the answer

Basically : \(\cfrac{1}{\lambda} - \cfrac{1}{\lambda + 1} = \cfrac{1}{\lambda (\lambda + 1)}\)

Answer 11

What I did was :

First simplify the given function : f(x)

It comes out to be :

\[\sum_{\lambda = 1}^{n} (x - \cfrac{1}{\lambda} )(x - \cfrac{1}{\lambda + 1} ) = \sum_{\lambda = 1}^{n} \cfrac{(\lambda x - 1)(\lambda x + x - 1)}{\lambda (\lambda + 1)}\]

Answer 12

And thus,

as \[f(0)= \sum_{\lambda = 1} ^{n} \cfrac{1}{\lambda (\lambda + 1)}\]