Is there any way to show that $\sum_{n=2}^{\infty}\frac{1}{n^2-1}$ converges without integral test?

Question

In my real analysis class, we are trying to show that $$\left(\frac{2 \times 2}{1 \times 3}\right)\left(\frac{4 \times 4}{3 \times 5}\right) \left(\frac{6 \times 6}{5 \times 7}\right)...$$ converges. I re-expressed this into the infinite product $(1 - \frac{1}{n^{2}-1})$. If I can show that the series of $\frac{1}{n^2 -1}$ converges then the proof will be complete, however we have yet to develop integration, and as a result is there any way to prove that this series goes to $0$.

Other methods to show that this product converges are also well appreciated!

Edit: Thanks so much! I completely forgot about the limit comparison test, been a while since I took calculus.

Answer 1

You can see that it increases.
On the other hand, when you group them so that a typical factor is $2\times4\over3\times3$, that sequence decreases.
So two sequences, one increasing, one decreasing. They are within $n\over n+1$ of each other, so yours is increasing and bounded above.

Answer 2

Since$$\lim_{n\to\infty}\frac{\frac1{n(n-1)}}{\frac1{n^2-1}}=1\neq0,$$and since the series$$\sum_{n=2}^\infty\frac1{n(n-1)}\left(=\sum_{n=2}^\infty\frac1{n-1}-\frac1n\right)$$converges, the series$$\sum_{n=2}^\infty\frac1{n^2-1}$$converges too.

Answer 3

We have $\frac{1}{n^2-1} \leq \frac{2}{n^2}$ for $n\geq2$. This can easily be verified. It follows that if $\sum_{n=1}^\infty\frac{1}{n^2}$ converges, then so does $\sum_{n=2}^\infty\frac{1}{n^2-1}$.

We can write

$$\sum\limits_{n=1}^\infty \frac{1}{n^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots$$

Also, we have

\begin{eqnarray*} 1 &\leq& 1 \\ \frac{1}{2^2} + \frac{1}{3^2} &\leq& \frac{1}{2^2} + \frac{1}{2^2} = \frac{1}{2} \\ \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} &\leq& \frac{4}{4^2} = \frac{1}{4} \end{eqnarray*} and so on.

Adding up all the terms on the left and the right we get $$\sum\limits_{n=1}^\infty\frac{1}{n^2} \leq 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \sum\limits_{n=0}^\infty \left(\frac{1}{2}\right)^n$$

This is a geometric series and it converges. Hence, $\sum_{n=1}^\infty\frac{1}{n^2}$ converges and, by implication, so does the starting series.

Answer 4

You can even calculate the exact value of $\sum_{n=2}^\infty \frac{1}{n^2 - 1}$. Namely, $$\frac{1}{n^2 - 1} = \frac{1}{2} \left( \frac{1}{n-1} - \frac{1}{n+1} \right) = \frac{1}{2} \left( \frac{1}{n-1} + \frac{1}{n} \right) - \frac{1}{2} \left( \frac{1}{n} + \frac{1}{n+1} \right).$$ Therefore, the series is a telescoping series, and $$\sum_{n=2}^N \frac{1}{n^2-1} = \frac{3}{4} - \frac{1}{2} \left( \frac{1}{N} + \frac{1}{N+1} \right).$$ Taking the limit as $N \to \infty$, it is then easy to see that $\sum_{n=2}^\infty \frac{1}{n^2-1} = \frac{3}{4}$.