How to sum the series $\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}$ I have been tinkering with telescoping sum and I've encountered by chance this series:
$$\sum_{n=4}^{+\infty}\dfrac{1}{(2n-7)(2n-6)}=\log(2)$$
I use Wolfram to sum this series because I am not sure if it is a telescoping sum. I wonder how can you sum this series? By converting the infinite sum to an improper integral?
I integrate this function:
$$\int_{4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}dx=\dfrac{\log(2)}{2}$$
I am new to improper integration and numerical infinite series. I thought the sum of the series and the result of the integral should match, but they don't.
It seems that I miss something very basic and crucial. I am learning myself Cal 3 so I may miss some crucial theorems. What are conditions for the sum of the series and the limit of the integral to be the same?
Come to think of it, summation is just a discrete version of integration, so the values naturally cannot be the same.
Is there a way to sum this series and obtain its sum? This doesn't seem to be a telescopic series in my humble opinion.
 A: The long and winding road. We will prove the result, only using that $H_N = \log N +\gamma + o(1)$, where $H_N$ is the Harmonic sum and $\gamma$ the Euler constant.

First, perform a change of index to get something more reasonable:
$$
\sum_{n=4}^{\infty}\frac{1}{(2n-7)(2n-6)}
= \sum_{k=0}^{\infty}\frac{1}{(2k+1)(2k+2)}
$$
Here, you may recognize that, for $k\geq 0$,
$$
\frac{1}{(2k+1)(2k+2)} = \frac{1}{2k+1}-\frac{1}{2k+2}
$$
and now you can use the following:
$$
\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)
= \sum_{k=0}^{n}\left(\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}\right)
= \sum_{k=1}^{2n+2}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k}
= H_{2n+2}-H_n
$$
Using now the fact that $H_{N} = \log N + \gamma +o(1)$, you get
$$
\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)
= H_{2n+2}-H_n
 = \log(2n+2) - \log n + o(1) = \log 2 + o(1)
$$
since $\log(2n+2) = \log 2 + \log (n+1) = \log 2 + \log n + o(1)$.
Taking $n\to \infty$, you get
$$
\sum_{k=4}^{\infty}\frac{1}{(2k-7)(2k-6)}
= \sum_{k=0}^{\infty}\frac{1}{(2k+1)(2k+2)} = \lim_{n\to\infty}\sum_{k=0}^{n}\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right) = \log 2
$$
as desired.
A: $$\frac{1}{(2n-6)(2n-7)}=\frac{1}{2n-7}-\frac{1}{ 2n-6  }$$
so your series is the Taylor series for $\log(1+x)=\sum_{k \ge 1} (-1)^{k-1} x^k/k$ evaluated at $x=1$.
A: $S=\displaystyle \sum_{n=4}^\infty \frac{1}{(2n-7)(2n-6)}=\sum_{n=4}^\infty\left(\int_{0}^1x^{2n-8}-x(x^{2n-8}) \mathrm{d}x\right)$
$=\displaystyle\int_{0}^1 \left(\sum_{n=4}^\infty x^{2n-8}-x(x^{2n-8}) \right)\mathrm{d}x$
$\text{(Note: the interchange of the summation is legitimate since the limits of integration are from 0 to 1)}$
$$S = \int_{0}^{1} \frac{1-x}{1-x^2} \mathrm{d}x=\log2 $$
A: Firstly, note that
$$\sum_{n=4}^{\infty}\dfrac{1}{(2n-7)(2n-6)}=\sum_{n=1}^\infty\dfrac{1}{(2n-1)(2n)}$$
Also, we can use partial fraction decomposition to find that
$$\frac{1}{2n(2n-1)}=\frac{1}{2n-1}-\frac{1}{2n}$$
So our sum is equivalent to
$$\sum_{n=1}^\infty\left(\frac{1}{2n-1}-\frac{1}{2n}\right)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$
But recall that the Maclaurin series for $\ln(1+x)$ which converges for $-1<x\leqslant1$ is
$$x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
So simply plugging in $x=1$ we have
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots=\ln(1+1)=\ln 2$$
as required.

I hope that was helpful. If you have any questions please don't hesitate to ask :)
A: You can use partial fractions. $\frac{1}{(2n-7)(2n-6)} = \frac{A}{2n-7} + \frac{B}{2n-6}$
If you combine you get A(2n-6) + B(2n-7) = 1, which must be true for all n>3.
When n = 4, 2A + B = 1
when n = 5, 4A + 3B = 1
Solving, we find A = 1 and B = -1, so
$\frac{1}{(2n-7)(2n-6)} = \frac{1}{2n-7} - \frac{1}{2n-6}$
which is an alternating series, in fact the alternating harmonic series.  I have memorized the fact that this series sums to ln(2), and I can prove it converges by the Alternating Series Test, but I don't recall offhand how to prove that it has that particular limit. In general, only telescoping series and geometric series can be summed exactly without advanced techniques that are not taught in calculus classes.
