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

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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!

For example

Also, how would you find a general equation for

Thanks in advance!

##### 2 Answers

#### 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

To further simplify, we use partial fraction decomposition on

Add up the right side:

Equate numerators:

Find

So,

Rewrite the series with decomposed fractions. Factor out

Now we can denote the sequence of partial sums by

So, let's examine the overall cancellation pattern. Be careful. So,

In general, we see that nearly every instance of

The partial sum is then

**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.**

#### Explanation:

Given:

#1/(n^2+5n+4)#

Note that:

#1/(n+1)-1/(n+4) = ((n+4)-(n+1))/((n+1)(n+4)) = 3/(n^2+5n+4)#

So:

#1/(n^2+5n+4) = 1/3(1/(n+1)-1/(n+4))#

So:

#3sum_(n=1)^N 1/(n^2+5n+4)#

#= sum_(n=1)^N (1/(n+1)-1/(n+4))#

#= sum_(n=1)^N 1/(n+1)- sum_(n=1)^N 1/(n+4)#

#= sum_(n=1)^N 1/(n+1)- sum_(n=4)^(N+3) 1/(n+1)#

#= sum_(n=1)^3 1/(n+1)+color(red)(cancel(color(black)(sum_(n=4)^N 1/(n+1))))-color(red)(cancel(color(black)(sum_(n=4)^N 1/(n+1))))-sum_(n=N+1)^(N+3) 1/(n+1)#

#= sum_(n=1)^3 1/(n+1)-sum_(n=N+1)^(N+3) 1/(n+1)#

#= 1/2+1/3+1/4-1/(N+2)-1/(N+3)-1/(N+4)#

#= 13/12-1/(N+2)-1/(N+3)-1/(N+4)#

So dividing both ends by

#sum_(n=1)^N 1/(n^2+5n+4) = 13/36-1/(3(N+2))-1/(3(N+3))-1/(3(N+4))#

So in the limit:

#sum_(n=1)^oo 1/(n^2+5n+4) = lim_(N->oo) sum_(n=1)^N 1/(n^2+5n+4)#

#color(white)(sum_(n=1)^oo 1/(n^2+5n+4)) = lim_(N->oo) (13/36-1/(3(N+2))-1/(3(N+3))-1/(3(N+4)))#

#color(white)(sum_(n=1)^oo 1/(n^2+5n+4)) = 13/36#

The more general case of