Show that series is convergent and less than 1

Question

Show that the series

$$ \sum_{k=1}^{\infty} \dfrac{2}{(k+1)(2k+1)} $$

is convergent and converges to a number less than $1$.

I tried the quotient test for series, but it didn't work. I don't know what else to do. Any hint would be very appreciated. Thanks!

Answer 1

Using comparsong test with $a_k=1/k^2$ you get that $$\lim_{k\to\infty} \frac{\frac{2}{(k+1)(2k+1)}}{1/k^2} = \lim_{k\to\infty}\frac{2k^2}{2k^2+3k+1}=1.$$ So both series are convergent.

Now, taking into account that $2k+1\ge k+1$ for all $k\ge 1$, you have $$\sum_{k=1}^\infty\frac{2}{(k+1)(2k+1)} \le \sum_{k=1}^\infty\frac{1}{(k+1)^2} =\sum_{k=1}^\infty\frac{1}{k^2}-1=\frac{\pi^2}{6}-1<1$$

Answer 2

Note that $\frac{2}{(k+1)(2k+1)}<\frac{1}{k(k+1)} = \frac{1}{k}-\frac {1}{k+1}$. But if we sum this from $1$ to $\infty$, all terms telescope except $1$ so the final sum is $1$. So we conclude that the original sum has a convergent value less than $1$.

Answer 3

$$\dfrac S4=-1+\dfrac12\sum_{r=0}\left(\dfrac1{2r+1}-\dfrac1{2r+2}\right)$$

$$=\ln(1+1)-\dfrac12$$