# How do you find the sum of an infinite sequence determined by a fraction with a polynomial?

## For example sum_(n=1)^∞1/(n^2+5n+4) Also, how would you find a general equation for sum_(n=1)^∞1/(n^2+xn+y)? Thanks in advance!

Mar 29, 2018

${s}_{N} = \frac{13}{36} - \frac{1}{3 \left(N + 4\right)}$

#### Explanation:

These are Telescoping Series , that is, when you write the sequence of partial sums, terms will cancel each other out in a particular pattern, yielding you a general equation for the sums.

Rewrite $\frac{1}{{n}^{2} + 5 n + 4}$ as $\frac{1}{\left(n + 1\right) \left(n + 4\right)}$

To further simplify, we use partial fraction decomposition on $\frac{1}{\left(n + 1\right) \left(n + 4\right)}$:

$\frac{1}{\left(n + 1\right) \left(n + 4\right)} = \frac{A}{n + 1} + \frac{B}{n + 4}$

Add up the right side:

$\frac{1}{\left(n + 1\right) \left(n + 4\right)} = \frac{A \left(n + 4\right) + B \left(n + 1\right)}{\left(n + 1\right) \left(n + 4\right)}$

Equate numerators:

$1 = A \left(n + 4\right) + B \left(n + 1\right)$

Find $A , B$ by choosing specific values of $n$ that will cause one term to cancel out and let you solve for the other:

$n = - 1 :$

$1 = 3 A , A = \frac{1}{3}$

$n = - 4 :$

$1 = - 3 B , B = - \frac{1}{3}$

So,

$\frac{1}{\left(n + 1\right) \left(n + 4\right)} = \frac{\frac{1}{3}}{n + 1} - \frac{\frac{1}{3}}{n + 4}$

Rewrite the series with decomposed fractions. Factor out $\frac{1}{3.}$

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

Now we can denote the sequence of partial sums by ${s}_{N}$, where $N$ is the $N t h$ partial sum.

s_N=1/3[(1/2-cancel(1/5))+(1/3cancel(-1/6))+(1/4cancel(-1/7))+(cancel(1/5)cancel(-1/8))+(cancel(1/6)cancel(-1/9))+(cancel(1/7)cancel(-1/10))+...+(1/(N+1)-1/(N+4))

So, let's examine the overall cancellation pattern. Be careful. So, $\frac{1}{2}$ never cancels out. It's a permanent part of the partial sum. The same goes for $\frac{1}{3} , \frac{1}{4.}$ The terms that do cancel out with a term farther down the sum have been cancelled above, even if the term they cancel with hasn't been written down (because it would be too far down the list).

In general, we see that nearly every instance of $\frac{1}{N + 1}$ cancels out with a term farther down the list:

${s}_{N} = \frac{1}{3} \left[\left(\frac{1}{2} - \cancel{\frac{1}{5}}\right) + \left(\frac{1}{3} \cancel{- \frac{1}{6}}\right) + \left(\frac{1}{4} \cancel{- \frac{1}{7}}\right) + \left(\cancel{\frac{1}{5}} - \frac{1}{8}\right) + \left(\cancel{\frac{1}{6}} - \frac{1}{9}\right) + \left(\cancel{\frac{1}{7}} - \frac{1}{10}\right) + \ldots + \cancel{\left(\frac{1}{N + 1}\right)} - \frac{1}{N + 4}\right)$

The partial sum is then

${s}_{N} = \frac{1}{3} \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{N + 4}\right) = \frac{1}{3} \left(\frac{13}{12} - \frac{1}{N + 4}\right) = \frac{13}{36} - \frac{1}{3 \left(N + 4\right)}$.

To find a general formula for the second series, factor the denominator, decompose the partial fraction, write some partial sums to test the cancellation pattern, and write the final sum with the relevant terms cancelled out as I have.

Mar 30, 2018

${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 5 n + 4} = \frac{13}{36}$

#### Explanation:

Given:

$\frac{1}{{n}^{2} + 5 n + 4}$

Note that:

$\frac{1}{n + 1} - \frac{1}{n + 4} = \frac{\left(n + 4\right) - \left(n + 1\right)}{\left(n + 1\right) \left(n + 4\right)} = \frac{3}{{n}^{2} + 5 n + 4}$

So:

$\frac{1}{{n}^{2} + 5 n + 4} = \frac{1}{3} \left(\frac{1}{n + 1} - \frac{1}{n + 4}\right)$

So:

$3 {\sum}_{n = 1}^{N} \frac{1}{{n}^{2} + 5 n + 4}$

$= {\sum}_{n = 1}^{N} \left(\frac{1}{n + 1} - \frac{1}{n + 4}\right)$

$= {\sum}_{n = 1}^{N} \frac{1}{n + 1} - {\sum}_{n = 1}^{N} \frac{1}{n + 4}$

$= {\sum}_{n = 1}^{N} \frac{1}{n + 1} - {\sum}_{n = 4}^{N + 3} \frac{1}{n + 1}$

$= {\sum}_{n = 1}^{3} \frac{1}{n + 1} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 4}^{N} \frac{1}{n + 1}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 4}^{N} \frac{1}{n + 1}}}} - {\sum}_{n = N + 1}^{N + 3} \frac{1}{n + 1}$

$= {\sum}_{n = 1}^{3} \frac{1}{n + 1} - {\sum}_{n = N + 1}^{N + 3} \frac{1}{n + 1}$

$= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{N + 2} - \frac{1}{N + 3} - \frac{1}{N + 4}$

$= \frac{13}{12} - \frac{1}{N + 2} - \frac{1}{N + 3} - \frac{1}{N + 4}$

So dividing both ends by $3$ we find:

${\sum}_{n = 1}^{N} \frac{1}{{n}^{2} + 5 n + 4} = \frac{13}{36} - \frac{1}{3 \left(N + 2\right)} - \frac{1}{3 \left(N + 3\right)} - \frac{1}{3 \left(N + 4\right)}$

So in the limit:

${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 5 n + 4} = {\lim}_{N \to \infty} {\sum}_{n = 1}^{N} \frac{1}{{n}^{2} + 5 n + 4}$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 5 n + 4}} = {\lim}_{N \to \infty} \left(\frac{13}{36} - \frac{1}{3 \left(N + 2\right)} - \frac{1}{3 \left(N + 3\right)} - \frac{1}{3 \left(N + 4\right)}\right)$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + 5 n + 4}} = \frac{13}{36}$

The more general case of ${\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} + x n + y}$ can be solved in a similar way if and only if ${x}^{2} + x n + y = \left(n + a\right) \left(n + b\right)$ for distinct integers $a$ and $b$.