Find the value of $S$ in term of $k$ (telescoping sums)

Question

Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.

Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.

I've tried to write the terms of $S$ in the following way:

\begin{align} \frac{1275}{1276} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1275\times1276} \\ \frac{1276}{1277} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{1276\times1277} \\ & \,\,\,\vdots \\ \frac{2548}{2549} & =\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{2548\times2549} \end{align}

But it's just not enough, because in $k$ we don't have $\frac{1}{2\times3}$, $\frac{1}{4\times5}$, etc.

Can you please help me solve the problem? Thanks!

Answer 1

Your sums are $k_{1275}$ and $S_{1275}$ where $$\cases{k_n=\sum_{i=1}^n\frac1{(2i-1)(2i)}\\S_n=\sum_{i=n}^{2n-2}\frac{i}{i+1}}$$

You can prove using induction that $$\forall n\geq 2,S_n=n-1+\frac1{2n}-k_n.$$

Answer 2

Note that your sum is $$\sum_{n=1}^{1275}{\frac{1}{2n(2n-1)}}$$ We can rewrite this using partial fractions as: $$\sum_{n=1}^{1275}{\bigg[\frac{1}{2n-1}-\frac{1}{2n}\bigg]}$$ and use telescoping from there.

Answer 3

$k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550} = \frac{1}{1}-\frac{1}{2} + \frac{1}{3}-\frac{1}{4} + ... + \frac{1}{2549}-\frac{1}{2550}$

Collecting even and odd and with some manipulation, we get

$k = H_{2550} - H_{1275}$ where $H_n$ = harmonic number

$S = 1 - \frac{1}{1276} + 1 - \frac{1}{1277} + ... + 1 - \frac{1}{2549}$

Collecting the ones and with some manipulation we get:

$S = 1274 - (H_{2550} - H_{1275}) + \frac{1}{2550} = 1274 - k + \frac{1}{2550}$