telescoping series (1 Viewer)

Lith_30 · May 12, 2022

$\\LHS=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\\\=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{2(k-1)(2k-3)}+\frac{1}{2k(2k-1)}\\\\=\frac{k(2k-1)}{2k(2k-1)}+\frac{\frac{1}{6}k(2k-1)}{2k(2k-1)}+\frac{\frac{1}{15}k(2k-1)}{2k(2k-1)}+...+\frac{\frac{1}{(k-1)(2k-3)}k(2k-1)}{2k(2k-1)}+\frac{1}{2k(2k-1)}\\\\=\frac{k(2k-1)(1+\frac{1}{6}+\frac{1}{15}+...+\frac{1}{(k-1)(2k-1)})+1}{2k(2k-1)}\\\\>\frac{k(2k-1)+1}{2k(2k-1)}\\\\>\frac{(2k-1)+1}{2k(2k-1)}\\\\>\frac{2k}{2k(2k-1)}\\\\>\frac{k}{k(2k-1)}\\\\>\frac{k}{(2k+1)(2k+2)}\\\\>RHS$

tbh I feel like there is a much more nicer way to solve this question, but I guess this method would work.

=)(= · May 12, 2022

I am not quite sure as to the 3 line

Lith_30 · May 12, 2022

=)(= said:
I am not quite sure as to the 3 line

Basically what I am doing is making all the fractions have the same denominator which is $2k(k-1)$

So for example $\frac{1}{2}$ would become $\frac{k(k-1)}{2k(k-1)}$ by multiplying numerator and denominator by k(k-1).

another example say for $\frac{1}{30}$ the working would be
$\\\frac{1}{30}\times\frac{k(k-1)}{k(k-1)}\\\\=\frac{1}{2}\times\frac{1}{15}\times\frac{k(k-1)}{k(k-1)}\\\\=\frac{\frac{1}{15}k(k-1)}{2k(k-1)}$

Trebla · May 12, 2022

Assuming k is a positive integer, another rather crude way to approach it is to realise that

$\dfrac{k}{(2k+1)(2k+2)}=\dfrac{1}{2}\times\dfrac{k}{(2k+1)(k+1)}$

Notice that

$\dfrac{k}{(2k+1)(k+1)} < 1$

because each factor in the denominator is clearly larger than the numerator, hence the whole denominator is less than the numerator so then

$\dfrac{k}{(2k+1)(2k+2)} < \dfrac{1}{2}$

Now notice that each bracketed component on the LHS is positive i.e. for positive integer k

$\dfrac{1}{2k-1}-\dfrac{1}{2k} > 0$

We can then easily argue that

$\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{2k-1}-\dfrac{1}{2k}\right) > \left(1-\dfrac{1}{2}\right)$

Combining the inequalities, we have the result

$\dfrac{k}{(2k+1)(2k+2)} < \dfrac{1}{2}< \left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{2k-1}-\dfrac{1}{2k}\right)$

Hence, the given inequality holds.

=)(= · May 13, 2022

ohh makes sense now thanks guys

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