# Calculating limit of series.

Given a series $$\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}$$, how do I find the limit? I understand I need to find the sequence of partial sums, which goes something like this $$s_n=\{\frac{5}{24}, \frac{7}{30}, \frac{27}{112}....\}$$which will probably converge at 0.25, but I am having trouble finding a description of the sequence on which I could evaluate the limit.

• partial fractions & telescope ? – Donald Splutterwit Nov 26 '20 at 16:38

The partial fraction decomposition of the summand is $$\frac1{2(2n-1)}-\frac1{2(2n+1)}-\frac1{4n}+\frac1{4(n+1)}$$ The first and second parts, and the third and fourth, telescope. In the infinite sum all parts except the $$n=1$$ instances of the first and third parts will cancel, so the infinite sum is $$\frac1{2(2×1-1)}-\frac1{4×1}=\frac14$$

• When you say they telescope, is there a general approach to solving these or do you need to look at some of the terms to see that they start cancelling after some $n$? – smejak Nov 26 '20 at 17:42
• @smejak "looking at some of the terms to see thst they start cancelling" is the general approach. – Parcly Taxel Nov 26 '20 at 17:45
• Alright, thank you. I just wasn't sure whether I don't need to use some proof technique to prove that all the terms actually cancel as $n\rightarrow\infty$. – smejak Nov 26 '20 at 17:51

As a hint:$$\sum_{n=1}^{\infty}\frac{4n+1}{2n(2n-1)(2n+1)(2n+2)}=\\ \sum_{n=1}^{\infty}\frac{4n+1}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\ \sum_{n=1}^{\infty}\frac{(2n-1)+(2n+2)}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\ \sum_{n=1}^{\infty}\frac{(2n-1)}{(2n-1)(2n+0)(2n+1)(2n+2)}+\sum_{n=1}^{\infty}\frac{(2n+2)}{(2n-1)(2n+0)(2n+1)(2n+2)}=\\ \sum_{n=1}^{\infty}\frac{1}{(2n+0)(2n+1)(2n+2)}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+0)(2n+1)}=\\$$ and now you have two telscopic series.can you take over now ?

Let $$\dfrac{4n+1}{(2n-1)2n(2n+1)(2n+2)}=f(n-1)-f(n)$$ where $$f(r)=\dfrac{ar+b}{(2r+1)(2r+2)}$$

so that $$\sum_{n=1}^m\dfrac{4n+1}{(2n-1)2n(2n+1)(2n+2)}=\sum_{n=1}^m(f(n-1)-f(n))=f(0)-f(m)$$

We need
$$f(n-1)-f(n)=\dfrac{(2n+1)(2n+2)(an-a+b)-(an+b)(2n-1)2n}{(2n-1)2n(2n+1)(2n+2)}$$

$$\implies4n+1=n^2(4a)+2n(?)+2b-2a$$

Comparing the coefficients of $$n^2,4a=0\iff a=?$$

Comparing the constants, $$2b-2a=1\implies b=\dfrac12$$

These values of $$a,b$$ satisfy the coefficients of $$n$$

Now set $$m\to\infty$$ and $$\lim_{m\to\infty}f(m)=?$$