-
Looking for HSC notes and resources? Check out our Notes & Resources page
telescoping series (1 Viewer)
- Thread starter =)(=
- Start date
Basically what I am doing is making all the fractions have the same denominator which is
So for example
another example say for
- Joined
- Feb 16, 2005
- Messages
- 8,392
- Gender
- Male
- HSC
- 2006
Assuming k is a positive integer, another rather crude way to approach it is to realise that
(2k+2)}=\dfrac{1}{2}\times\dfrac{k}{(2k+1)(k+1)})
Notice that
(k+1)} < 1)
because each factor in the denominator is clearly larger than the numerator, hence the whole denominator is less than the numerator so then
(2k+2)} < \dfrac{1}{2})
Now notice that each bracketed component on the LHS is positive i.e. for positive integer k
We can then easily argue that
+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{2k-1}-\dfrac{1}{2k}\right) > \left(1-\dfrac{1}{2}\right))
Combining the inequalities, we have the result
(2k+2)} < \dfrac{1}{2}< \left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{2k-1}-\dfrac{1}{2k}\right))
Hence, the given inequality holds.
Notice that
because each factor in the denominator is clearly larger than the numerator, hence the whole denominator is less than the numerator so then
Now notice that each bracketed component on the LHS is positive i.e. for positive integer k
We can then easily argue that
Combining the inequalities, we have the result
Hence, the given inequality holds.