In which interval is the value of $\sum\limits_{k=1}^\infty \frac{1}{{\left(k + 3\right)}^{2}} $ The question is

But none of these possible answers makes any sense to me.
I know that $\displaystyle\sum_{k=1}^\infty \left( \displaystyle\frac{1}{{\left(k + 3\right)}^{2}} \right)$ is an over estimate of $\displaystyle\int_{1}^\infty \displaystyle\frac{1}{{\left(x + 3\right)}^{2}} \, dx = \displaystyle\frac{1}{4}$ 
So I think the value should be in the interval of $[1/16,L]$ where $L > 1/4$ but none of these answers reflect this.
 A: By the integral test we have for a monotonically decreasing non-negative function $f$:
\begin{align*}
\int_1^\infty f(x)dx\leq\sum_{k=1}^\infty f(k)\leq f(1)+\int_1^\infty f(x)dx.
\end{align*}
In your case, take $f(x)=1/(x+3)^2$. Then this gives
\begin{align*}
\frac{1}{4}\leq\sum_{k=1}^\infty\frac{1}{(k+3)^2}\leq \frac{1}{16}+\frac{1}{4}=\frac{5}{16},
\end{align*}
so answer number 2 is correct.
A: We can actually find the value of this series:
Note that:
$$\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \dots + \frac{1}{n^2} = \frac{\pi^2}{6}$$
Now, take a look at the actual sum. Let $S$ be the value of our sum:
$$S = \sum_{k = 1}^{\infty} \frac{1}{(k+3)^2} = \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots$$
Next, if we add to both sides the first original $3$ terms we get:
$$S + \frac{1}{1} + \frac{1}{4} + \frac{1}{9} = \frac{1}{1} + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots = \frac{\pi^2}{6}$$
Now we simply subtract to get the value of $S$:
$$S = \frac{\pi^2}{6} - \left(\frac{1}{1} + \frac{1}{4} + \frac{1}{9}\right) \approx 0.2838$$
Looking at the answers, we see that it is: The second choice
