# Does this series converge or diverge...?

I need to determine if this series converges:

$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$ I tried to solve this using two methods.
Method 1
This is positive, continuous and monotonically decreasing, so I used the integral test first. $$\int_{1}^\infty \frac{1}{2n(2n+1)}$$ = $$ln|\sqrt{2x + 1} + \sqrt{2x} |_1 ^\infty$$ which is $$\infty$$. And therefore the series diverges.
Method 2
I performed the comparison test with $$\zeta (2)$$. As $$\frac{1}{n^2} \gt \frac{1}{2n(2n+1)}$$ for all $$n \in \mathbb N$$, and $$\frac{1}{n^2}$$ converges to $$\frac{\pi^2}{6}$$, the given series converges as well.
Which method is wrong?
Thank you!

• Your integral calculation is wrong. The series converges as you've shown in your second method.
– Snaw
Commented Dec 12, 2021 at 11:34
• Your integration is not correct. The series converges. Commented Dec 12, 2021 at 11:34
• I am so sorry, I made a calculation error.Thanks everyone. Commented Dec 12, 2021 at 11:41

The first approach has an error. It turns out that$$\int\frac{\mathrm dx}{2x(2x+1)}=\log\left(\sqrt{\frac x{2x+1}}\right).$$Since$$\lim_{x\to\infty}\log\left(\sqrt{\frac x{2x+1}}\right)=\log\left(\sqrt{\frac12}\right),$$the integral converges.
• Your integral is wrong, it should be $$\int_1^\infty\frac{1}{2x(2x+1)} \,dx=\frac{1}{2}(\log(x)-\log(2x+1))\Big|_1^\infty$$
• Also, note that $$\frac{1}{2n(2n+1)}=\frac{1}{2n}-\frac{1}{2n+1}$$