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# Find a general formula for  S(n)=sum_4^(n)(1/(k-3)-1/(k)) and evaluate the limit S=lim_(n rarr oo) S(n)?

Oct 3, 2017

${\sum}_{4}^{n} \left(\frac{1}{k - 3} - \frac{1}{k}\right) = \frac{\left(n - 3\right) \left(11 {n}^{2} - 18 n + 4\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

${\sum}_{4}^{\infty} \left(\frac{1}{k - 3} - \frac{1}{k}\right) = \frac{11}{6}$

#### Explanation:

We seek:

$S = {\sum}_{4}^{\infty} \left(\frac{1}{k - 3} - \frac{1}{k}\right)$

Firstly, let us find a formula for:

$S \left(n\right) = {\sum}_{4}^{n} \left(\frac{1}{k - 3} - \frac{1}{k}\right)$

This a a typical summation of differences, and as such many terms will cancel. If we expand the summation this becomes clear:

$S \left(n\right) = \left(\frac{1}{1} - \textcolor{b l u e}{\frac{1}{4}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{2} - \textcolor{g r e e n}{\frac{1}{5}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 5\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{3} - \textcolor{red}{\frac{1}{6}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 6\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{b l u e}{\frac{1}{4}} - \frac{1}{7}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{g r e e n}{\frac{1}{5}} - \frac{1}{8}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 8\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{red}{\frac{1}{6}} - \frac{1}{9}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 9\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{n - 6} - \textcolor{p u r p \le}{\frac{1}{n - 3}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 3\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{n - 5} - \frac{1}{n - 2}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 2\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{n - 4} - \frac{1}{n - 1}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 1\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{p u r p \le}{\frac{1}{n - 3}} - \frac{1}{n}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n\right)$

And now if we cancel the terms which appear as positive from one term and negative and another:

$S \left(n\right) = \left(\frac{1}{1} - \textcolor{b l u e}{\cancel{\frac{1}{4}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 4\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{2} - \textcolor{g r e e n}{\cancel{\frac{1}{5}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 5\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{3} - \textcolor{red}{\cancel{\frac{1}{6}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 6\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{b l u e}{\cancel{\frac{1}{4}}} - \cancel{\frac{1}{7}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{g r e e n}{\cancel{\frac{1}{5}}} - \cancel{\frac{1}{8}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 8\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{red}{\cancel{\frac{1}{6}}} - \cancel{\frac{1}{9}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = 9\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\cancel{\frac{1}{n - 6}} - \textcolor{p u r p \le}{\cancel{\frac{1}{n - 3}}}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 3\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\cancel{\frac{1}{n - 5}} - \frac{1}{n - 2}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 2\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\cancel{\frac{1}{n - 4}} - \frac{1}{n - 1}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n - 1\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\textcolor{p u r p \le}{\cancel{\frac{1}{n - 3}}} - \frac{1}{n}\right) + \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(k = n\right)$

Which leaves us with a much reduced collection of terms:

$S \left(n\right) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{11}{6} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}$ ..... [A]

Although not strictly required, We can simplify this result (this is typically the case in an exam question when $S \left(n\right)$ is sought prior to finding the limiting case $S \left(\infty\right)$)

$\setminus \setminus = \frac{11}{6} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}$

$\setminus \setminus = \frac{11 n \left(n - 1\right) \left(n - 2\right) - 6 n \left(n - 1\right) - 6 n \left(n - 2\right) - 6 \left(n - 1\right) \left(n - 2\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{11 n \left(n - 1\right) \left(n - 2\right) - 6 n \left(n - 1\right) - 6 n \left(n - 2\right) - 6 \left(n - 1\right) \left(n - 2\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{11 {n}^{3} - 33 {n}^{2} + 22 n - 6 {n}^{2} + 6 n - 6 {n}^{2} + 12 n - 6 {n}^{2} + 18 n - 12}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{11 {n}^{3} - 51 {n}^{2} + 58 n - 12}{6 n \left(n - 1\right) \left(n - 2\right)}$

$\setminus \setminus = \frac{\left(n - 3\right) \left(11 {n}^{2} - 18 n + 4\right)}{6 n \left(n - 1\right) \left(n - 2\right)}$

Factorization was performed via a calculator at the last step, the answer at [A] would suffice

So finally, we seek:

$S = {\lim}_{n \rightarrow \infty} S \left(n\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \left\{\frac{11}{6} - \frac{1}{n - 2} - \frac{1}{n - 1} - \frac{1}{n}\right\} \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ (from [A])
$\setminus \setminus = \frac{11}{6}$

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Validation:

It is always worth checking a few sums just to check the result is consistent

By direct summation
S(4) = 3/4
S(5) = 21/21
S(5) = 73/60

And using the derived summation formula gives the same result

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