Question

Evaluate:

[tex]\displaystyle\sf{\sum_{n=1}^{\infty}\dfrac{1}{n^2+5n+6}}​

Answer 1

ANSWER:

To Evaluate:

$\:\:\:\bullet\:\:\:\displaystyle\sf{\sum_{n=1}^{\infty}\dfrac{1}{n^2+5n+6}}$

Solution:

We are given that,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{1}{n^2+5n+6}}$

Let, n² + 5n + 6 be p(n).

So,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{1}{p(n)}}$

So, first we will solve the p(n), and then proceed further.

Hence,

⇒ p(n) = n² + 5n + 6

So,

⇒ p(n) = n² + 5n + 6

Splitting the middle term,

⇒ p(n) = n² + 2n + 3n + 6

Taking common,

⇒ p(n) = n(n + 2) + 3(n+ 2)

So,

⇒ p(n) = (n + 2)(n + 3)

We were given,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{1}{p(n)}}$

So, putting value of p(n),

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{1}{(n+2)(n+3)}}$

Now, we need to make changes with the numerator so that, we simplify the expression.

We know that, 1 = 3 - 2.

(we took this, because denominator had 3 and 2.)

So,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{3-2}{(n+2)(n+3)}}$

To simplify it more, we'll add and subtract 'n' in the numerator.

So,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{3-2+n-n}{(n+2)(n+3)}}$

On rearranging and grouping,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{(n+3)-(n+2)}{(n+2)(n+3)}}$

Now, we'll split the fraction,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{(n+3)}{(n+2)(n+3)}-\dfrac{(n+2)}{(n+2)(n+3)}}$

On simplifying,

$\displaystyle\implies\sf{\sum_{n=1}^{\infty}\dfrac{1}{(n+2)}-\dfrac{1}{(n+3)}}$

Now, we will, open the summation, and place values of n from 1 to infinite.

So,

$\displaystyle\implies\sf{\left(\dfrac{1}{1+2}-\dfrac{1}{1+3}\right)+\left(\dfrac{1}{2+2}-\dfrac{1}{2+3}\right)+\dots\dots\dots+\left(\dfrac{1}{\infty+2}-\dfrac{1}{\infty+3}\right)}$

Now,

$\displaystyle\implies\sf{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dots\dots\dots+\dfrac{1}{\infty}-\dfrac{1}{\infty}}$

We can see that every term starting from 1/4 gets cancelled.

So,

$\displaystyle\implies\sf{\dfrac{1}{3}}$

Hence,

$\displaystyle\implies\bf{\sum_{n=1}^{\infty}\dfrac{1}{n^2+5n+6}=\dfrac{1}{3}}$