Find $ \frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1} $

Question

Find the following sum

$$ \frac{1}{2^2 –1} + \frac{1}{4^2 –1} + \frac{1}{6^2 –1} + \ldots + \frac{1}{20^2 –1} $$

I am not able to find any short trick for it. Is there any short trick or do we have to simplify and add it?

Answer 1

$$\sum_{n=1}^{10}\frac{1}{4n^2-1}=\frac12\sum_{n=1}^{10}\frac{1}{2n-1}-\frac{1}{2n+1}=\frac12\left(1-\frac{1}{21}\right)=\frac{10}{21}$$

Answer 2

Lemma to find the sum of $n$ terms of a series each term of which is composed of the reciprocal of the product of $r$ factors in arithmetical progression, the first factors of the several terms being in the same arithmetical progression:

Write down the $nth$ term, strike off a factor from the beginning, divide by the number of factors so diminished and by the common difference, change the sign and add a constant.

In our case, the general term is $T_n$=$$\frac 1 {(2n)^2-1}$$

Which is simply, $$\frac 1 {(2n-1)(2n+1)}$$

Applying the lemma, we get:

$$S_n=C-(\frac 1 2)(\frac 1 {2n+1})$$

$$S_n=C-\frac 1 {4n+2}$$ Where $C$ is a constant quantity.

Putting $n=1$ in general term, we get:

$$S_1=\frac 1 3=C- \frac 1 {4+2}$$

That gives $$C=\frac 1 2$$

Thus, $$S_n=\frac 1 2-\frac 1 {4n+2}$$

Putting $n=10$ gives:

$$S_{10}=\frac 1 2-\frac 1 {40+2}$$ $$S_{10}=\frac 1 2-\frac 1 {42}$$ $$Or, S_{10}=\frac {20} {42}=\frac {10} {21}$$
Which is the desired answer.

The method is a bit, long... But it is great for the so-called "bashing" through such problems. If you wish to see a proof, I refer you to Hall and Knight's Higher Algebra.

(P.S. I did this long answer with this specific method just so I could get a hang of the typing system in StackExchange. So, please forgive me even if the method seems useless... I'm new here...)

Answer 3

Alternatively to the telescoping sum decomposition, there is an easy pattern

$$\frac13$$

$$\frac13+\frac1{15}=\frac25$$

$$\frac13+\frac1{15}+\frac1{35}=\frac37$$

$$\frac13+\frac1{15}+\frac1{35}+\frac1{63}=\frac49$$

$$\cdots$$